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: //Created by Aminul on 20/12/21.
  
  import java.io.*;
  import java.util.*;
  
  import static java.lang.Math.*;
  
  public class D {
      public static void main(String[] args) throws Exception {
          Scanner in = new Scanner(System.in);
          PrintWriter pw = new PrintWriter(System.out);
          int test = in.nextInt();
          for (int t = 1; t <= test; t++) {
              int m = in.nextInt(), n = in.nextInt();
              int[][] arr = new int[m][n];
              for (int i = 0; i < m; i++) {
                  for (int j = 0; j < n; j++) {
                      arr[i][j] = in.nextInt();
                  }
              }
  
              pw.println(binarySearch(m, n, arr));
          }
  
          pw.close();
      }
  
      static int binarySearch(int m, int n, int[][] arr) {
          int l = 0, r = (int) 1e9;
          int res = 0;
          while (l <= r) {
              int mid = l + ((r - l) >> 1);
              if (check(m, n, arr, mid)) {
                  res = mid;
                  l = mid + 1;
              } else {
                  r = mid - 1;
              }
          }
          return res;
      }
  
      static boolean check(int m, int n, int[][] arr, int min) {
          int[] freq = new int[m];
          int[] found = new int[n];
          for (int i = 0; i < n; i++) {
              for (int j = 0; j < m; j++) {
                  if (arr[j][i] >= min) {
                      freq[j]++;
                      found[i]++;
                  }
              }
          }
  
          for (int i = 0; i < n; i++) {
              if (found[i] == 0) {
                  return false;
              }
          }
  
          for (int i = 0; i < m; i++) {
              if (freq[i] > 1) {
                  return true;
              }
          }
  
          return false;
      }
  
      static void debug(Object... obj) {
          System.err.println(Arrays.deepToString(obj));
      }
  }