description: Casimir has a rectangular piece of paper with a checkered field of size n × m. Initially, all cells of the field are white.
  
  Let us denote the cell with coordinates i vertically and j horizontally by (i, j). The upper left cell will be referred to as (1, 1) and the lower right cell as (n, m).
  
  Casimir draws ticks of different sizes on the field. A tick of size d (d > 0) with its center in cell (i, j) is drawn as follows: 
  
    1. First, the center cell (i, j) is painted black. 
    2. Then exactly d cells on the top-left diagonally to the center and exactly d cells on the top-right diagonally to the center are also painted black. 
    3. That is all the cells with coordinates (i - h, j ± h) for all h between 0 and d are painted. In particular, a tick consists of 2d + 1 black cells. 
  
  
  
  An already painted cell will remain black if painted again. Below you can find an example of the 4 × 9 box, with two ticks of sizes 2 and 3.
  
  <image>
  
  You are given a description of a checkered field of size n × m. Casimir claims that this field came about after he drew some (possibly 0) ticks on it. The ticks could be of different sizes, but the size of each tick is at least k (that is, d ≥ k for all the ticks).
  
  Determine whether this field can indeed be obtained by drawing some (possibly none) ticks of sizes d ≥ k or not.
  
  Input
  
  The first line contains an integer t (1 ≤ t ≤ 100) — the number test cases.
  
  The following lines contain the descriptions of the test cases. 
  
  The first line of the test case description contains the integers n, m, and k (1 ≤ k ≤ n ≤ 10; 1 ≤ m ≤ 19) — the field size and the minimum size of the ticks that Casimir drew. The following n lines describe the field: each line consists of m characters either being '.' if the corresponding cell is not yet painted or '*' otherwise.
  
  Output
  
  Print t lines, each line containing the answer to the corresponding test case. The answer to a test case should be YES if the given field can be obtained by drawing ticks of at least the given size and NO otherwise.
  
  You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes, and YES will all be recognized as positive answers).
  
  Example
  
  Input
  
  
  8
  2 3 1
  *.*
  ...
  4 9 2
  *.*.*...*
  .*.*...*.
  ..*.*.*..
  .....*...
  4 4 1
  *.*.
  ****
  .**.
  ....
  5 5 1
  .....
  *...*
  .*.*.
  ..*.*
  ...*.
  5 5 2
  .....
  *...*
  .*.*.
  ..*.*
  ...*.
  4 7 1
  *.....*
  .....*.
  ..*.*..
  ...*...
  3 3 1
  ***
  ***
  ***
  3 5 1
  *...*
  .***.
  .**..
  
  
  Output
  
  
  NO
  YES
  YES
  YES
  NO
  NO
  NO
  NO
  
  Note
  
  The first sample test case consists of two asterisks neither of which can be independent ticks since ticks of size 0 don't exist.
  
  The second sample test case is already described in the statement (check the picture in the statement). This field can be obtained by drawing ticks of sizes 2 and 3, as shown in the figure.
  
  The field in the third sample test case corresponds to three ticks of size 1. Their center cells are marked with \color{blue}{blue}, \color{red}{red} and \color{green}{green} colors: *.*.  
  ---  
  *\color{blue}{*}**  
  .\color{green}{*}\color{red}{*}.  
  ....  
    
  The field in the fourth sample test case could have been obtained by drawing two ticks of sizes 1 and 2. Their vertices are marked below with \color{blue}{blue} and \color{red}{red} colors respectively: .....  
  ---  
  *...*  
  .*.*.  
  ..\color{red}{*}.*  
  ...\color{blue}{*}.  
    
  The field in the fifth sample test case can not be obtained because k = 2, and the last asterisk in the fourth row from the top with coordinates (4, 5) can only be a part of a tick of size 1.
  
  The field in the sixth sample test case can not be obtained because the top left asterisk (1, 1) can't be an independent tick, since the sizes of the ticks must be positive, and cannot be part of a tick with the center cell in the last row, since it is separated from it by a gap (a point, '.') in (2, 2).
  
  In the seventh sample test case, similarly, the field can not be obtained by the described process because the asterisks with coordinates (1, 2) (second cell in the first row), (3, 1) and (3, 3) (leftmost and rightmost cells in the bottom) can not be parts of any ticks.
solution: import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  
  import static java.lang.Math.*;
  
  public class Main {
  
      //------------------------------------------CONSTANTS---------------------------------------------------------------
      public static final int MOD = (int) (1e9 + 7);
  
      //---------------------------------------------I/0------------------------------------------------------------------
  
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader()
          {
              br = new BufferedReader(
                      new InputStreamReader(System.in));
          }
  
          String next()
          {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  }
                  catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() { return Integer.parseInt(next()); }
  
          long nextLong() { return Long.parseLong(next()); }
  
          double nextDouble()
          {
              return Double.parseDouble(next());
          }
  
          String nextLine()
          {
              String str = "";
              try {
                  str = br.readLine();
              }
              catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  
      public static void print(String s) {
          System.out.println(s);
      }
  
      public static void main(String[] args) {
          FastReader sc = new FastReader();
  
          int t = 1;
          t = sc.nextInt();
  
          for(int i = 0; i<t; i++) {
              System.out.println(solve(sc));
          }
      }
  
      //------------------------------------------------SOLVE-------------------------------------------------------------
  
      public static String solve(FastReader sc) {
          int n = sc.nextInt(), m = sc.nextInt(), k = sc.nextInt();
  
          int[][] mx = new int[n][m];
          int[][] mx2 = new int[n][m];
          for (int i = 0; i<n; i++) {
              String s = sc.next();
              for (int j = 0; j<m; j++) {
                  mx[i][j] = s.charAt(j) == '*'?1:0;
              }
          }
  
          for (int i = 1; i<n; i++) {
              for (int j = 1; j<m; j++) {
                  if (mx[i][j] == 1) {
                      int d = 0;
                      int newI = i;
                      int jDis = 1;
                      while (newI-1 >= 0 && j-jDis >= 0 && j+jDis < m) {
                          if (mx[newI-1][j-jDis] == 1 && mx[newI-1][j+jDis] == 1) {
                              d++;
                          } else {
                              break;
                          }
                          newI--;
                          jDis++;
                      }
                      if (d >= k) {
                          mx2[i][j] = 1;
                          newI = i;
                          jDis = 1;
                          int times = 0;
                          while (times < d) {
                              mx2[newI-1][j-jDis] = 1;
                              mx2[newI-1][j+jDis] = 1;
                              newI--; jDis++;
                              times++;
                          }
                      }
                  }
              }
          }
  
          for (int i = 0; i<n; i++) {
              for (int j = 0; j<m; j++) {
                  if (mx[i][j] != mx2[i][j]) {
                      return "NO";
                  }
              }
          }
          return "YES";
      }
  
      public static boolean isOrdered(int[] a) {
          for (int i = 0; i<a.length-1; i++) {
              if (a[i] > a[i+1])
                  return false;
          }
  
          return true;
      }
  
      public static void cycleShift(int[] a, int l, int r, int d) {
          for (int i = 0; i<d; i++) {
              int al = a[l];
              for (int j = l; j<r-1; j++) {
                  a[j] = a[j+1];
              }
              a[r-1] = al;
          }
      }
  
      //-----------------------------------------------FUNCTIONS----------------------------------------------------------
  
      public static String printArray(int[] a) {
          StringBuilder r = new StringBuilder("");
  
          for(int i = 0; i<a.length; i++) {
              r.append(a[i] + " ");
          }
          return r.toString();
      }
  
      // first element that is not less than target
      public static int lowerBound(int[] arr, int begin, int end, int target) {
          while(begin < end) {
              int mid = begin + (end - begin) / 2;
              // When the element is less than the mid target
              if(arr[mid] < target)
                  // begin to mid + 1, arr [begin] value is less than or equal target
                  begin = mid + 1;
                  // When the mid target element when greater than or equal
              else if(arr[mid] >= target)
                  end = mid;
          }
          return begin;
      }
  
      // first element that is greater than target
      public static int upperBound(int[] arr, int begin, int end, int target) {
          while(begin < end) {
              int mid = begin + (end - begin) / 2;
              if(arr[mid] <= target)
                  begin = mid + 1;
              else
                  end = mid;
          }
          return begin;
      }
  
      // generate permutations (call search and access through permutations)
      public static class Permutations {
          List<List<Integer>> permutations = new ArrayList<>();
          List<Integer> permutation = new ArrayList<>();
          boolean[] taken;
          int[] a;
  
          public Permutations(int[] a) {
              this.a = a;
              taken = new boolean[a.length];
          }
  
          void search() {
              if (permutation.size()==a.length) {
                  permutations.add(new ArrayList<>(permutation));
              } else {
                  for (int i = 0; i<a.length; i++) {
                      if (taken[i]) {
                          continue;
                      }
                      taken[i] = true;
                      permutation.add(a[i]);
                      search();
                      taken[i] = false;
                      permutation.remove(permutation.size()-1);
                  }
              }
          }
      }
  
      // generate subsets (call search and access subsets)
      public static class Subsets {
          List<List<Integer>> subsets = new ArrayList<>();
          List<Integer> subset = new ArrayList<>();
  
          void search(int[] a, int x) {
              if (x == a.length) {
                  subsets.add(new ArrayList<>(subset));
              } else {
                  subset.add(a[x]);
                  search(a, x+1);
                  subset.remove(subset.size() - 1);
                  search(a, x+1);
              }
          }
      }
  
      public static int gcd(int a, int b) {
          while (b != 0) {
              int t = b;
              b = a % b;
              a = t;
          }
          return a;
      }
  
      public static int lcm(int a, int b) {
          return (a*b)/gcd(a, b);
      }
  
      public static int[] prefixSum(int[] a) {
          int n = a.length;
          int[] ps = new int[n];
          ps[0] = a[0];
  
          for (int i = 1; i<n; i++) {
              ps[i] = ps[i-1] + a[i];
          }
  
          return ps;
      }
  
      public static int[][] prefixSum2D(int[][] a) {
          int n = a.length;
          int m = a[0].length;
          int[][] ps = new int[n][m];
          ps[0][0] = a[0][0];
  
          for (int j = 1; j<m; j++) {
              ps[0][j] = ps[0][j-1] + a[0][j];
          }
          for (int i = 1; i<n; i++) {
              ps[i][0] = ps[i-1][0] + a[i][0];
          }
  
          for (int i = 1; i<n; i++) {
              for (int j = 1; j<m; j++) {
                  ps[i][j] = ps[i-1][j] + ps[i][j-1] - ps[i-1][j-1] + a[i][j];
              }
          }
  
          return ps;
      }
  
      public static double log2(double a) {
          return log(a)/log(2);
      }
  
      //----------------------------------------DATA-STRUCTURES-----------------------------------------------------------
  
      // Range queries (i.e. RMQ) in O(1), others with functions in O(log n)
      public static class SparseTable {
          int[][] st;
          int n;
          int k;
  
          public SparseTable(int[] a) {
              n = a.length;
              k = (int) log2(n);
              st = new int[n][k+1];
  
              for (int i = 0; i<n; i++) {
                  st[i][0] = a[i];
              }
  
              // 1<<j == 2^j
              for (int j = 1; j<=k; j++) {
                  for (int i = 0; i + (1 << j) <= n; i++) {
                      st[i][j] = min(st[i][j-1], st[i + (1 << (j - 1))][j - 1]);
                  }
              }
          }
  
          //not always available O(log n) -- change the constructor to use it
          public long sumQuery(int l, int r) {
              long sum = 0;
              for (int j = k; j>=0; j--) {
                  if ((1 << j) <= r - l + 1) {
                      sum += st[l][j];
                      l += 1 << j;
                  }
              }
  
              return sum;
          }
  
          //always available O(1) -- change this and the constructor for rangeMaxQuery
          public int rangeMinQuery(int l, int r) {
              int j = (int) log2(r - l + 1);
              int min = min(st[l][j], st[r - (1 << j) + 1][j]);
              return min;
          }
  
      }
  
      //Range Sum Dynamic Queries O(log n)
      public static class BinaryIndexedTree {
  
          int[] bit;
  
          public BinaryIndexedTree(int[] a) {
              int n = a.length;
              bit = new int[n+1];
              // Store the actual values in BITree[]
              // using update()
              for(int i = 0; i < n; i++)
                  updateBit(i, a[i]);
          }
  
          // O(log n) update delta at index in the array (do it on the array and then call this function)
          public void updateBit(int index, int delta) {
              index += 1;
  
              while (index <= bit.length) {
                  bit[index] += delta;
  
                  index += index & (-index);
              }
          }
  
          // Computes sum a[0...index] in O(log n)
          public int getSum(int index) {
              int sum = 0;
  
              index += 1;
  
              while (index > 0) {
                  sum += bit[index];
  
                  index -= index & (-index);
              }
  
              return sum;
          }
      }
  
      public static class UnionFind {
  
          int[] link;
          int[] size;
  
          UnionFind(int n) {
              link = new int[n];
              size = new int[n];
              for (int i = 0; i<n; i++) {
                  link[i] = i;
                  size[i] = 1;
              }
          }
  
          int find(int x) {
              while (x != link[x]) x = link[x];
              return x;
          }
  
          boolean same(int a, int b) {
              return find(a) == find(b);
          }
  
          void unite(int a, int b) {
              a = find(a);
              b = find(b);
  
              if (size[a] < size[b]) {
                  int temp = a;
                  a = b;
                  b = temp;
              }
  
              size[a] += size[b];
              link[b] = a;
          }
      }
  
      public static class SegmentTree {
  
          int st[];
  
          public SegmentTree(int arr[]) {
              int n = arr.length;
  
              int x = (int) (Math.ceil(log2(n)));
  
              int maxSize = 2 * (int) Math.pow(2, x) - 1;
  
              st = new int[maxSize];
  
              build(arr, 0, n-1, 0);
          }
  
          int build(int arr[], int ss, int se, int si) {
              if (ss == se) {
                  st[si] = arr[ss];
                  return arr[ss];
              }
  
              int mid = ss + (se - ss) / 2; //ss+se/2
  
              st[si] = build(arr, ss, mid, si * 2 + 1) +
                      build(arr, mid + 1, se, si * 2 + 2);
              return st[si];
          }
  
          private int getSumUtil(int ss, int se, int l, int r, int si) {
              if (l <= ss && r >= se) {
                  return st[si];
              }
  
              if (se < l || ss > r) {
                  return 0;
              }
  
              int mid = ss + (se - ss) / 2;
              return getSumUtil(ss, mid, l, r, 2 * si + 1) +
                      getSumUtil(mid + 1, se, l, r, 2 * si + 2);
          }
  
          void updateValueUtil(int ss, int se, int i, int diff, int si) {
              if (i < ss || i > se) {
                  return;
              }
  
              st[si] = st[si] + diff;
              if (se != ss) {
                  int mid = ss + (se - ss) / 2;
                  updateValueUtil(ss, mid, i, diff, 2 * si + 1);
                  updateValueUtil(mid + 1, se, i, diff, 2 * si + 2);
              }
          }
  
          int getSum(int n, int l, int r) {
              if (l < 0 || r > n - 1 || l > r) {
                  return -1;
              }
  
              return getSumUtil(0, n-1, l, r, 0);
          }
  
          // don't do it directly on the array but use this function that will
          // do everything for you
          void updateValue(int arr[], int n, int i, int newValue) {
              if (i < 0 || i > n - 1) {
                  return;
              }
  
              int diff = newValue - arr[i];
  
              arr[i] = newValue;
  
              updateValueUtil(0, n - 1, i, diff, 0);
          }
      }
  
      // Similar to Pair in c++, non-null objects as parameters
      public static class Pair<T extends Comparable<T>, Q extends Comparable<Q>> implements Comparable<Pair<T, Q>>{
          T a;
          Q b;
  
          public Pair(T a, Q b) {
              this.a = a;
              this.b = b;
          }
  
          @Override
          public boolean equals(Object o) {
              if (this == o) return true;
              if (o == null || getClass() != o.getClass()) return false;
              Pair<?, ?> pair = (Pair<?, ?>) o;
              return a.equals(pair.a) && b.equals(pair.b);
          }
  
          @Override
          public int hashCode() {
              return Objects.hash(a, b);
          }
  
  
          @Override
          public int compareTo(Pair<T, Q> tqPair) {
              int compareA = this.a.compareTo(tqPair.a);
  
              if (compareA == 0) {
                  return this.b.compareTo(tqPair.b);
              } else {
                  return compareA;
              }
          }
      }
  
  
  }