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 test {
      public static void main(String[] args) throws IOException {
          FastScanner fs=new FastScanner();
          PrintWriter out = new PrintWriter(System.out);
          int T = fs.nextInt();
          for (int tt=0; tt<T; tt++) {
              int m = fs.nextInt();
              int n = fs.nextInt();
  
              int[][] mat = new int[m][n];
              for (int i=0; i<m; i++) {
                  mat[i] = fs.readArray(n);
              }
              int res = 0;
  
              int[] best = new int[n];
              for (int i=0; i<m; i++) {
                  for (int j=0; j<n; j++) {
                      best[j] = Math.max(best[j], mat[i][j]);
                  }
              }
              res = Integer.MAX_VALUE;
              for (int i=0; i<n; i++) {
                  res = Math.min(res, best[i]);
              }
              int res2 = res;
              if (m<=n-1) {
              }
              else {
                  res = 0;
                  int[][] topTwo = new int[m][2];
  
                  for (int i=0; i<m; i++) {
                      for (int j=0; j<n; j++) {
                          if (topTwo[i][0]<=mat[i][j]) {
                              topTwo[i][1] = topTwo[i][0];
                              topTwo[i][0] = mat[i][j];
                          }
                          else if (topTwo[i][1]<mat[i][j]) {
                              topTwo[i][1] = mat[i][j];
                          }
                      }
                  }
                  for (int i=0; i<m; i++) {
                      // System.out.println(Arrays.toString(topTwo[i]));
                      res = Math.max(res, topTwo[i][1]);
                  }
                  res = Math.min(res,res2);
              }
  
              
              out.println(res);
          	
  
          }
          out.close();
  
      }
  
      static final Random random=new Random();
  
      static void ruffleSort(int[] a) {
          int n=a.length;//shuffle, then sort
          for (int i=0; i<n; i++) {
              int oi=random.nextInt(n), temp=a[oi];
              a[oi]=a[i]; a[i]=temp;
          }
          Arrays.sort(a);
      }
  
      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());
          }
          int[] readArray(int n) {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  
  }
  