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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
   
  public class taskd {
      public static void main(String[] args) {
          FastScanner in = new FastScanner();
          PrintWriter out = new PrintWriter(System.out);
          int tt = in.nextInt();
          while (tt-- > 0) {
              taskd sol = new taskd();
              sol.solve(in, out);
          }
          out.flush();
      }
  
      void solve(FastScanner in, PrintWriter out) {
          int m = in.nextInt(), n = in.nextInt();
          long a[][] = new long[m][n];
          for (int i = 0; i < m; i++) {
              for (int j = 0; j < n; j++) {
                  a[i][j] = in.nextLong();
              }
          }
          long low = 1, high = (long) 1e9;
          long ans = 1;
          while (low <= high) {
              long mid = (low + high) >> 1;
              boolean every_col = true;
              for (int j = 0; j < n; j++) {
                  boolean has = false;
                  for (int i = 0; i < m; i++) {
                      has |= a[i][j] >= mid;
                  }
                  every_col &= has;
              }
              boolean dup = false;
              for (int i = 0; i < m; i++) {
                  int cnt = 0;
                  for (int j = 0; j < n; j++) {
                      cnt += (a[i][j] >= mid ? 1 : 0);
                  }
                  dup |= cnt >= 2;
              }
              if (every_col && dup) {
                  ans = mid;
                  low = mid + 1;
              } else {
                  high = mid - 1;
              }
          }
          out.println(ans);
      }
  
      static class FastScanner {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer("");
          String next() {
              while (!st.hasMoreTokens()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
          int nextInt() { return Integer.parseInt(next()); }
          long nextLong() { return Long.parseLong(next()); }
          float nextFloat() { return Float.parseFloat(next()); }
          double nextDouble() { return Double.parseDouble(next()); }
      }
  }