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.HashSet;
  import java.util.StringTokenizer;
  
  /**
   * Accomplished using the EduTools plugin by JetBrains https://plugins.jetbrains.com/plugin/10081-edutools
   */
  
  
  public class Main {
      public static void main(String[] args) {
          InputStream inputStream = System.in;
          OutputStream outputStream = System.out;
          InputReader in = new InputReader(inputStream);
          PrintWriter out = new PrintWriter(outputStream);
          Task solver = new Task();
          solver.solve(1, in, out);
          out.close();
      }
  
      public static class Task {
  
          public void solve(int cnt, InputReader in, PrintWriter out) {
              cnt = in.nextInt();
              for (int cc = 0; cc < cnt; cc++) {
                  int m = in.nextInt();
                  int n = in.nextInt();
                  int maxMin = Integer.MAX_VALUE;
                  int[][] joy = new int[m][n];
                  //每个朋友最大的joy，和对应商店
                  int[][] maxJoy = new int[n][2];
                  for (int i = 0; i < m; i++) {
                      for (int j = 0; j < n; j++) {
                          joy[i][j] = in.nextInt();
                          if (joy[i][j] >= maxJoy[j][0]) {
                              maxJoy[j][0] = joy[i][j];
                              maxJoy[j][1] = i;
                          }
                      }
                  }
                  //遍历每个朋友最大joy对应的商店，因为最多选n-1家，所以如果各自不同，就要抛弃一家
                  HashSet<Integer> shops = new HashSet<>();
                  for (int i = 0; i < n; i++) {
                      shops.add(maxJoy[i][1]);
                      maxMin = Math.min(maxJoy[i][0], maxMin);
                  }
                  if (shops.size() >= n) {
                      //抛弃一家
                      //找到每一行第二大的
                      int ret = 0;
                      for(int i=0;i<m;i++) {
                          for(int j=0;j<n;j++) {
                              Arrays.sort(joy[i]);
                              ret = Math.max(ret, joy[i][n-2]);
                          }
                      }
                      maxMin = Math.min(ret, maxMin);
                  }
                  out.println(maxMin);
              }
          }
      }
  
      static class InputReader {
          public BufferedReader reader;
          public StringTokenizer tokenizer;
  
          public InputReader(InputStream stream) {
              reader = new BufferedReader(new InputStreamReader(stream), 32768);
              tokenizer = null;
          }
  
          public String next() {
              while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                  try {
                      tokenizer = new StringTokenizer(reader.readLine());
                  } catch (IOException e) {
                      throw new RuntimeException(e);
                  }
              }
              return tokenizer.nextToken();
          }
  
          public int nextInt() {
              return Integer.parseInt(next());
          }
  
          public long nextLong() {
              return Long.parseLong(next());
          }
  
      }
  }