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.*;
  
  public class codeforces_762_D {
      private static void solve(FastIOAdapter in, PrintWriter out) {
          int m = in.nextInt();
          int n = in.nextInt();
          int[][] mm = new int[m][];
          for (int i = 0; i < m; i++) {
              int[] c = in.readArray(n);
              mm[i] = c;
          }
          int[][][] minv = new int[n][m][2];
          for (int i = 0; i < n; i++) {
              for (int j = 0; j < m; j++) {
                  minv[i][j] = new int[]{mm[j][i], j};
              }
              Arrays.sort(minv[i], (i1, i2) -> {
                  if (i1[0] != i2[0]) return Integer.compare(i2[0], i1[0]);
                  else return Integer.compare(i1[1], i2[1]);
              });
          }
  
          int minM = Math.min(n - 1, m);
  
          HashSet<Integer> set = new HashSet<>();
  
          int min = Integer.MAX_VALUE;
          for (int i = 0; i < n; i++) {
              set.add(minv[i][0][1]);
              min = Math.min(min, minv[i][0][0]);
          }
          if (set.size() > minM) {
              int max = Integer.MIN_VALUE;
              for (int i = 0; i < n; i++) {
                  for (int j = 1; j < m; j++) {
                      int v = minv[i][j][0];
                      int place = minv[i][j][1];
                      if (set.contains(place)) {
                          max = Math.max(max, v);
                      }
                  }
              }
              int max2 = Integer.MIN_VALUE;
              for (int i = 0; i < m; i++) {
                  int cmax1 = Integer.MIN_VALUE;
                  int cmax2 = Integer.MIN_VALUE;
                  for (int j = 0; j < n; j++) {
                      if (mm[i][j] > cmax1) {
                          cmax2 = cmax1;
                          cmax1 = mm[i][j];
                      } else if (mm[i][j] > cmax2) {
                          cmax2 = mm[i][j];
                      }
                  }
                  max2 = Math.max(max2, cmax2);
              }
              min = Math.min(min, Math.max(max, max2));
          }
          out.println(min);
      }
  
      public static void main(String[] args) throws Exception {
          try (FastIOAdapter ioAdapter = new FastIOAdapter()) {
              int count = 1;
              count = ioAdapter.nextInt();
              while (count-- > 0) {
                  solve(ioAdapter, ioAdapter.out);
              }
          }
      }
  
      static void ruffleSort(int[] arr) {
          int n = arr.length;
          Random rnd = new Random();
          for (int i = 0; i < n; ++i) {
              int tmp = arr[i];
              int randomPos = i + rnd.nextInt(n - i);
              arr[i] = arr[randomPos];
              arr[randomPos] = tmp;
          }
          Arrays.sort(arr);
      }
  
      static class FastIOAdapter implements AutoCloseable {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          public PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter((System.out))));
          StringTokenizer st = new StringTokenizer("");
  
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          String nextLine() {
              try {
                  return br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
                  return null;
              }
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          int[] readArray(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++) a[i] = nextInt();
              return a;
          }
  
          long[] readArrayLong(int n) {
              long[] a = new long[n];
              for (int i = 0; i < n; i++) a[i] = nextLong();
              return a;
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          @Override
          public void close() throws Exception {
              out.flush();
              out.close();
              br.close();
          }
      }
  }
  