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.util.Scanner;
  import java.io.DataInputStream;
  import java.io.FileInputStream;
  
  import java.io.IOException;
  
  public class NewYearsProblem {
  
      static boolean check(Integer[][] a, int n, int m, int k) {
          boolean ch = false;
          for (int i = 0; i < n; i++) {
              int c = 0;
             
              for (int j = 0; j < m; j++) {
                  if(a[i][j]>=k){
                      ++c;
                  }
              }
              if(c>=2){
                 ch = true;
                 break;
              }
          }
          
          if(ch){
              int c = 0;
                  k:  for (int j = 0; j < m; j++) {
                      for (int l = 0; l < n; l++) {
                          if(a[l][j]>=k){
                              ++c;
                              if(c>=m){
                                  return true;
                              }
                             continue k;
                          }
                      }
                  }
                  return c>= m;
          }
          return false;
  
      }
   
      public static void main(String args[]) throws IOException {
          Reader scan = new Reader();
          int t = scan.nextInt();
          StringBuilder sb = new StringBuilder();
          while (t-- > 0) {
              int n = scan.nextInt();
              int m = scan.nextInt();
              Integer[][] a = new Integer[n][m];
              for (int i = 0; i < n; i++) {
                  for (int j = 0; j < m; j++) {
                      a[i][j] = scan.nextInt();
                  }
              }
              int l = 0;
              int r = 1000000001;
              int res = 0;
              while (l <= r) {
                  int mid = (l + r) / 2;
                  if (check(a, n, m, mid)) {
                      l = mid + 1;
                      res = Math.max(res, mid);
                  } else {
                      r = mid - 1;
                  }
              }
              sb.append(res+"\n");
          }
          System.out.println(sb);
      }
  }
  
  
  
   
  
  class Reader {
      
      final private int BUFFER_SIZE = 1 << 16;
      private DataInputStream din;
      private byte[] buffer;
      private int bufferPointer, bytesRead;
      
      public Reader() {
          din = new DataInputStream(System.in);
          buffer = new byte[BUFFER_SIZE];
          bufferPointer = bytesRead = 0;
      }
      
      public Reader(String file_name) throws IOException {
          din = new DataInputStream(
                  new FileInputStream(file_name));
          buffer = new byte[BUFFER_SIZE];
          bufferPointer = bytesRead = 0;
      }
      
      public String readLine() throws IOException {
          byte[] buf = new byte[64]; // line length
          int cnt = 0, c;
          while ((c = read()) != -1) {
              if (c == '\n') {
                  if (cnt != 0) {
                      break;
                  } else {
                      continue;
                  }
              }
              buf[cnt++] = (byte) c;
          }
          return new String(buf, 0, cnt);
      }
      
      public int nextInt() throws IOException {
          int ret = 0;
          byte c = read();
          while (c <= ' ') {
              c = read();
          }
          boolean neg = (c == '-');
          if (neg) {
              c = read();
          }
          do {
              ret = ret * 10 + c - '0';
          } while ((c = read()) >= '0' && c <= '9');
          
          if (neg) {
              return -ret;
          }
          return ret;
      }
      
      public long nextLong() throws IOException {
          long ret = 0;
          byte c = read();
          while (c <= ' ') {
              c = read();
          }
          boolean neg = (c == '-');
          if (neg) {
              c = read();
          }
          do {
              ret = ret * 10 + c - '0';
          } while ((c = read()) >= '0' && c <= '9');
          if (neg) {
              return -ret;
          }
          return ret;
      }
      
      public double nextDouble() throws IOException {
          double ret = 0, div = 1;
          byte c = read();
          while (c <= ' ') {
              c = read();
          }
          boolean neg = (c == '-');
          if (neg) {
              c = read();
          }
          
          do {
              ret = ret * 10 + c - '0';
          } while ((c = read()) >= '0' && c <= '9');
          
          if (c == '.') {
              while ((c = read()) >= '0' && c <= '9') {
                  ret += (c - '0') / (div *= 10);
              }
          }
          
          if (neg) {
              return -ret;
          }
          return ret;
      }
      
      private void fillBuffer() throws IOException {
          bytesRead = din.read(buffer, bufferPointer = 0,
                  BUFFER_SIZE);
          if (bytesRead == -1) {
              buffer[0] = -1;
          }
      }
      
      private byte read() throws IOException {
          if (bufferPointer == bytesRead) {
              fillBuffer();
          }
          return buffer[bufferPointer++];
      }
      
      public void close() throws IOException {
          if (din == null) {
              return;
          }
          din.close();
      }
  }
  