description: Vlad has n friends, for each of whom he wants to buy one gift for the New Year.
  
  There are m shops in the city, in each of which he can buy a gift for any of his friends. If the j-th friend (1 ≤ j ≤ n) receives a gift bought in the shop with the number i (1 ≤ i ≤ m), then the friend receives p_{ij} units of joy. The rectangular table p_{ij} is given in the input.
  
  Vlad has time to visit at most n-1 shops (where n is the number of friends). He chooses which shops he will visit and for which friends he will buy gifts in each of them.
  
  Let the j-th friend receive a_j units of joy from Vlad's gift. Let's find the value α=min\\{a_1, a_2, ..., a_n\}. Vlad's goal is to buy gifts so that the value of α is as large as possible. In other words, Vlad wants to maximize the minimum of the joys of his friends.
  
  For example, let m = 2, n = 2. Let the joy from the gifts that we can buy in the first shop: p_{11} = 1, p_{12}=2, in the second shop: p_{21} = 3, p_{22}=4.
  
  Then it is enough for Vlad to go only to the second shop and buy a gift for the first friend, bringing joy 3, and for the second — bringing joy 4. In this case, the value α will be equal to min\{3, 4\} = 3
  
  Help Vlad choose gifts for his friends so that the value of α is as high as possible. Please note that each friend must receive one gift. Vlad can visit at most n-1 shops (where n is the number of friends). In the shop, he can buy any number of gifts.
  
  Input
  
  The first line of the input contains an integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input.
  
  An empty line is written before each test case. Then there is a line containing integers m and n (2 ≤ n, 2 ≤ n ⋅ m ≤ 10^5) separated by a space — the number of shops and the number of friends, where n ⋅ m is the product of n and m.
  
  Then m lines follow, each containing n numbers. The number in the i-th row of the j-th column p_{ij} (1 ≤ p_{ij} ≤ 10^9) is the joy of the product intended for friend number j in shop number i.
  
  It is guaranteed that the sum of the values n ⋅ m over all test cases in the test does not exceed 10^5.
  
  Output
  
  Print t lines, each line must contain the answer to the corresponding test case — the maximum possible value of α, where α is the minimum of the joys from a gift for all of Vlad's friends.
  
  Example
  
  Input
  
  
  5
  
  2 2
  1 2
  3 4
  
  4 3
  1 3 1
  3 1 1
  1 2 2
  1 1 3
  
  2 3
  5 3 4
  2 5 1
  
  4 2
  7 9
  8 1
  9 6
  10 8
  
  2 4
  6 5 2 1
  7 9 7 2
  
  
  Output
  
  
  3
  2
  4
  8
  2
solution: import java.io.*;
  import java.util.Arrays;
  import java.util.Locale;
  import java.util.StringTokenizer;
  
  public class DDD {
  
      private final boolean MULTI_TEST = true;
      private final String FILE_NAME = "";
  
      private void solve() {
          int m = ri(), n = ri();
          int[][] p = new int[m][n];
          for(int i = 0; i < m; i++) {
              for(int j = 0; j < n; j++) {
                  p[i][j] = ri();
              }
          }
  
          int[] max = new int[n];
          for(int j = 0; j < n; j++) {
              int maximal = Integer.MIN_VALUE;
              for(int i = 0; i < m; i++) {
                  if(maximal < p[i][j]) maximal = p[i][j];
              }
              max[j] = maximal;
          }
  
          int min = Integer.MAX_VALUE;
          for(int i = 0; i < n; i++) {
              if(max[i] < min) min = max[i];
          }
  
          int maxSecond = Integer.MIN_VALUE;
          for(int i = 0; i < m; i++) {
              p[i] = sort(p[i]);
              if(maxSecond < p[i][n - 2]) {
                  maxSecond = p[i][n-2];
              }
          }
          out.println(Math.min(min, maxSecond));
      }
  
      //////////////////////////////////////////////////////////////////
  
      long sqrtLong(long x) {
          long root = (long) Math.sqrt(x);
          while (root * root > x) --root;
          while ((root + 1) * (root + 1) <= x) ++root;
          return root;
      }
  
      int gcd(int a, int b) {
          if (b == 0)
              return a;
          else
              return gcd(b, a % b);
      }
  
      int lcm(int a, int b) {
          return a / gcd(a, b) * b;
      }
  
      int phi(int n) {
          int result = n;
          for (int i = 2; i * i <= n; ++i)
              if (n % i == 0) {
                  while (n % i == 0)
                      n /= i;
                  result -= result / i;
              }
          if (n > 1)
              result -= result / n;
          return result;
      }
  
      int factmod(int n, int p) {
          int res = 1;
          while (n > 1) {
              res = (res * ((n / p) % 2 == 1 ? p - 1 : 1)) % p;
              for (int i = 2; i <= n % p; ++i)
                  res = (res * i) % p;
              n /= p;
          }
          return res % p;
      }
  
      int binpow(int a, int n) {
          int res = 1;
          while (n != 0) {
              if ((n & 1) != 0)
                  res *= a;
              a *= a;
              n >>= 1;
          }
          return res;
      }
  
      String solPyatnashki() {
          int a[] = new int[16];
          for (int i = 0; i < 16; ++i)
              a[i] = ri();
  
          int inv = 0;
          for (int i = 0; i < 16; ++i)
              if (a[i] != 0)
                  for (int j = 0; j < i; ++j)
                      if (a[j] > a[i])
                          ++inv;
          for (int i = 0; i < 16; ++i)
              if (a[i] == 0)
                  inv += 1 + i / 4;
  
  
          return ((inv & 1) != 0 ? "No Solution" : "Solution Exists");
      }
  
      boolean isPrime(int number) {
          for (int i = 2; i * i <= number; i++) {
              if (number % i == 0) {
                  return false;
              }
          }
          return number > 1;
      }
  
      boolean[] primes(int border) {
          boolean[] isPrimes = new boolean[border + 1];
          Arrays.fill(isPrimes, true);
          isPrimes[0] = false;
          isPrimes[1] = false;
          for (int i = 2; i < border + 1; i++) {
              if (!isPrimes[i]) continue;
              for (int k = i * 2; k < border; k += i) {
                  isPrimes[k] = false;
              }
          }
          return isPrimes;
      }
  
      int[] sort(int[] arr) {
          Integer[] a = new Integer[arr.length];
          for (int i = 0; i < arr.length; i++) {
              a[i] = arr[i];
          }
          Arrays.sort(a);
          for (int i = 0; i < a.length; i++) {
              arr[i] = a[i];
          }
          return arr;
      }
  
      long[] sort(long[] arr) {
          Long[] a = new Long[arr.length];
          for (int i = 0; i < arr.length; i++) {
              a[i] = arr[i];
          }
          Arrays.sort(a);
          for (int i = 0; i < a.length; i++) {
              arr[i] = a[i];
          }
          return arr;
      }
  
      int max(int[] a) {
          int max = Integer.MIN_VALUE;
          for (int i = 0; i < a.length; i++) {
              if (a[i] > max) max = a[i];
          }
          return max;
      }
  
      long fastMod(String s1, long n2) {
          long num = 0;
          for (int i = 0; i < s1.length() - 1; i++) {
              num += Integer.parseInt(String.valueOf(s1.charAt(i)));
              num *= 10;
              if (num >= n2) {
                  num = num % n2;
              }
          }
          return (num + Integer.parseInt(String.valueOf(s1.charAt(s1.length() - 1)))) % n2;
      }
  
      //////////////////////////////////////////////////////////////////
  
      private boolean yesNo(boolean yes) {
          return yesNo(yes, "YES", "NO");
      }
  
      private boolean yesNo(boolean yes, String yesString, String noString) {
          out.println(yes ? yesString : noString);
          return yes;
      }
  
      //////////////////////////////////////////////////////////////////
  
      private long rl() {
          return Long.parseLong(rs());
      }
  
      private double rd() {
          return Double.parseDouble(rs());
      }
  
      private int[] ria(int n) {
          int[] a = new int[n];
          for (int i = 0; i < n; ++i) a[i] = ri();
          return a;
      }
  
      private int ri() {
          return Integer.parseInt(rs());
      }
  
      private String rs() {
          while (!tok.hasMoreTokens()) {
              String nextLine = readLine();
              if (null == nextLine) return null;
              tok = new StringTokenizer(nextLine);
          }
  
          return tok.nextToken();
      }
  
      private String readLine() {
          try {
              return in.readLine();
          } catch (IOException e) {
              throw new RuntimeException(e);
          }
      }
  
      //////////////////////////////////////////////////////////////////
  
      private BufferedReader in;
      private StringTokenizer tok;
      private PrintWriter out;
  
      private void initFileIO(String inputFileName, String outputFileName) throws FileNotFoundException {
          in = new BufferedReader(new FileReader(inputFileName));
          out = new PrintWriter(outputFileName);
      }
  
      private void initConsoleIO() throws IOException {
          in = new BufferedReader(new InputStreamReader(System.in));
          out = new PrintWriter(System.out);
      }
  
      private void initIO() throws IOException {
          Locale.setDefault(Locale.US);
  
          if (!FILE_NAME.isEmpty()) {
              initFileIO(FILE_NAME + ".in", FILE_NAME + ".out");
          } else {
              if (new File("input.txt").exists()) {
                  initFileIO("input.txt", "output.txt");
              } else {
                  initConsoleIO();
              }
          }
  
          tok = new StringTokenizer("");
      }
  
      //////////////////////////////////////////////////////////////////
  
      private void run() {
          try {
              long timeStart = System.currentTimeMillis();
              initIO();
              if (MULTI_TEST) {
                  int t = ri();
                  while (t-- > 0) {
                      solve();
                      out.flush();
                  }
              } else
                  solve();
              out.close();
              long timeEnd = System.currentTimeMillis();
              System.err.println("Time(ms) = " + (timeEnd - timeStart));
          } catch (Exception e) {
              e.printStackTrace();
              throw new RuntimeException(e);
          }
      }
  
      public static void main(String[] args) {
          new DDD().run();
      }
  
      ///////////////////////////////////////////////////////////////////
  }
  