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 B {
      static int mod = (int) 1e9 + 7;
  
      public static void main(String[] args) throws IOException {
          BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
          PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
          int T = Integer.parseInt(bufferedReader.readLine());
          while (T-- > 0) {
              bufferedReader.readLine();
              StringTokenizer stringTokenizer = new StringTokenizer(bufferedReader.readLine());
              int m = Integer.parseInt(stringTokenizer.nextToken());
              int n = Integer.parseInt(stringTokenizer.nextToken());
  
              int[][] arr = new int[m][n];
              int max_val[] = new int[n];
              for (int i = 0; i < m; i++) {
                  stringTokenizer = new StringTokenizer(bufferedReader.readLine());
                  for (int j = 0; j < n; j++) {
                      arr[i][j] = Integer.parseInt(stringTokenizer.nextToken());
  
                      if (max_val[j] < arr[i][j])
                          max_val[j] = arr[i][j];
                  }
  
              }
              int ans=max_val[0];
              for (int i = 1; i < n; i++) {
                  ans=Math.min(max_val[i], ans);
              }
              if(m <= n-1){
                  out.println(ans);
                  continue;
              }
              int secLargest=0;
              for (int i = 0; i < m; i++) {
                  Arrays.sort(arr[i]);
  
                  secLargest=Math.max(secLargest, arr[i][arr[0].length-2]);
              }
              ans=Math.min(secLargest, ans);
              out.println(ans);
          }
          out.close();
      }
  
      public static long power(long a, int p) {
          long res = 1;
          while (p > 0) {
              if ((p & 1) == 1)
                  res = (res * a) % mod;
              a = (a * a) % mod;
              p >>= 1;
          }
          return res;
      }
  
      static long gcd(long a, long b) {
          if (a == 0)
              return b;
          return gcd(b % a, a);
      }
  
      static long lcm(int a, int b) {
          return (a / gcd(a, b)) * b;
      }
  }
  
  /*
   * boolean[] sieve = new boolean[1_000_001];
   * Arrays.fill(sieve, true);
   * sieve[0] = false;
   * sieve[1] = false;
   * 
   * for (int x = 2; x * x < sieve.length; x++) {
   * if (sieve[x]) {
   * for (int j = x; j * x < sieve.length; j++) {
   * sieve[j * x] = false;
   * }
   * }
   * }
   */
  