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.*;
  import java.io.*;
  
  public class NewYearsProblem{
     public static void main(String[] args) throws IOException{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        
        int t=Integer.parseInt(br.readLine());
        for(int j=0; j<t; j++){
           //get input
           br.readLine();
           
           StringTokenizer st=new StringTokenizer(br.readLine());
           int m=Integer.parseInt(st.nextToken());
           int n=Integer.parseInt(st.nextToken());
           
           long[][] joy=new long[m][n];
           for(int i=0; i<m; i++){
              StringTokenizer st2=new StringTokenizer(br.readLine());
              for(int k=0; k<n; k++) joy[i][k]=Long.parseLong(st2.nextToken());
           }
           
           //System.out.println(works(joy, 9));
           
           //binary search
           long lo=0;
           long hi=(long)Math.pow(10, 9);
           long ans=0;
           
           while(lo<=hi){
              long mid=(lo+hi)/2;
              if(works(joy, mid)){
                 ans=mid;
                 lo=mid+1;
              }
              else{
                 hi=mid-1;
              }
           }
           
           System.out.println(ans);
        }
     }
     static boolean works(long[][] joy, long min){
        for(int f=0; f<joy[0].length; f++){
           boolean works=false;
           for(int s=0; s<joy.length; s++){
              if(joy[s][f]>=min){
                 works=true;
                 break;
              }
           }
           if(!works) return false;
        }
        //Number of stores visited limitation
        boolean ok=false;
        for(int s=0; s<joy.length; s++){
           int cnt=0;
           for(int f=0; f<joy[0].length; f++){
              if(joy[s][f]>=min){
                 cnt++;
                 if(cnt>=2){
                    ok=true;
                    break;
                 }
              }
           }
        }
        if(!ok) return false;
        return true;
     }
  }