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: //package com.codeforces.Round762;
  
  import java.io.*;
  import java.util.Arrays;
  
  public class problem04 {
  
  
      /////////////////////////////////////////////////////////////////////////
      ///        How to Solve Problem in CP ?   ( if you feel problem is hard then your approach is wrong )
      // ==>> Step01 :- Understanding problem statement clearly. Then Only Move Forward
      //      Step02 :- Think a lot about Solution. if you are confident that your solution might be correct Then only Move Forward
      //      Step03 :- First think of brute force then move to optimal approach
      //      Step04 :- Finally Code ( there is no any sense to code. if you not follow about steps okk)
      ///////////////////////////////////////////////////////////////////////////
      public static void main(String[] args)throws IOException {
          Reader scan=new  Reader();
          BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
          int t=scan.nextInt();
          while (t-->0) {
               int m=scan.nextInt();
               int n=scan.nextInt();
               int[][] arr=new int[m][n];
               int[] c=new int[n];
               for (int i = 0; i < m; i++) {
                   for (int j = 0; j < n; j++) {
                       arr[i][j]=scan.nextInt();
                       c[j]=Math.max(c[j],arr[i][j]);
                   }
                   Arrays.sort(arr[i]);
               }
               int res=0;
              for (int i = 0; i < m; i++) {
                  res=Math.max(res,arr[i][n-2]);
              }
              for (int i = 0; i < n; i++) {
                  res=Math.min(res,c[i]);
              }
              bw.write(res+"");
              bw.newLine();
          }
          bw.flush();
      }
  
      //FAST READER
      static class Reader {
          public int BS = 1<<16;
          public char NC = (char)0;
          byte[] buf = new byte[BS];
          int bId = 0, size = 0;
          char c = NC;
          double num = 1;
          BufferedInputStream in;
  
          public Reader() {
              in = new BufferedInputStream(System.in, BS);
          }
  
          public Reader(String s) throws FileNotFoundException {
              in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
          }
  
          public char nextChar(){
              while(bId==size) {
                  try {
                      size = in.read(buf);
                  }catch(Exception e) {
                      return NC;
                  }
                  if(size==-1)return NC;
                  bId=0;
              }
              return (char)buf[bId++];
          }
  
          public int nextInt() {
              return (int)nextLong();
          }
  
          public long nextLong() {
              num=1;
              boolean neg = false;
              if(c==NC)c=nextChar();
              for(;(c<'0' || c>'9'); c = nextChar()) {
                  if(c=='-')neg=true;
              }
              long res = 0;
              for(; c>='0' && c <='9'; c=nextChar()) {
                  res = (res<<3)+(res<<1)+c-'0';
                  num*=10;
              }
              return neg?-res:res;
          }
  
          public double nextDouble() {
              double cur = nextLong();
              return c!='.' ? cur:cur+(cur < 0 ? -1*nextLong()/num : nextLong()/num);
          }
  
          public String next() {
              StringBuilder res = new StringBuilder();
              while(c<=32)c=nextChar();
              while(c>32) {
                  res.append(c);
                  c=nextChar();
              }
              return res.toString();
          }
      }
  
  }
  