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 E_Gardener_and_Tree implements Runnable{
  
         public static void main(String[] args) throws Exception {
          new Thread(null, new E_Gardener_and_Tree(), "E_Gardener_and_Tree", 1<<28).start();
         }
      public void run(){
           WRITE YOUR CODE HERE!!!!
           JUST WRITE EVERYTHING HERE WHICH YOU WRITE IN MAIN!!!
         }
  
    }
  */
  /////**************************************************************************
  
  
  public class D_New_Year_s_Problem{
      public static void main(String[] args) {
          FastScanner s= new FastScanner();
          //PrintWriter out=new PrintWriter(System.out);
          //end of program
          //out.println(answer);
          //out.close();
          StringBuilder res = new StringBuilder();
          int t=s.nextInt();
          int p=0;
          while(p<t){
          int n=s.nextInt();// for shops
          int m=s.nextInt();// for friends
          long matrix[][]= new long[n][m];
          long min=Long.MAX_VALUE;
          for(int i=0;i<n;i++){
              for(int j=0;j<m;j++){
                  matrix[i][j]=s.nextLong();
                  min=Math.min(min,matrix[i][j]);
              }
          }
          long start=min;
          long end=power(10,9)+10;
          long mid=0;
          while(start<end){
              mid=(start+end)/2;
              if(start+1==end){
                  break;
              }
              long array[]= new long[m];// fr firriend
              long yoyo=0;
              for(int i=0;i<n;i++){
                  long hh=0;
                  for(int j=0;j<m;j++){
                      long num=matrix[i][j];
                      if(num>=mid){
                          array[j]=1;
                          hh++;
                      }
                  }
                  yoyo=Math.max(yoyo,hh);
              }
              int check1=0;
              for(int i=0;i<m;i++){
                  if(array[i]==0){
                      check1=1;
                      break;
                  }
              }
              if(check1==1){
                  end=mid;
              }
              else{
                  if(yoyo>1){
                      start=mid;
                  }
                  else{
                      end=mid;
                  }
  
              }
  
          }
          res.append(start+" \n");
          p++;
      }
      System.out.println(res);
  
      }
      static class FastScanner {
  		BufferedReader br;
  		StringTokenizer st;
   
  		public FastScanner(String s) {
  			try {
  				br = new BufferedReader(new FileReader(s));
  			} catch (FileNotFoundException e) {
  				// TODO Auto-generated catch block
  				e.printStackTrace();
  			}
  		}
   
  		public FastScanner() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
   
  		String nextToken() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					// TODO Auto-generated catch block
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
   
  		int nextInt() {
  			return Integer.parseInt(nextToken());
  		}
   
  		long nextLong() {
  			return Long.parseLong(nextToken());
  		}
   
  		double nextDouble() {
  			return Double.parseDouble(nextToken());
  		}
  	}
  
  
      static long modpower(long x, long y, long p)
       {
        long res = 1; // Initialize result
     
        x = x % p; // Update x if it is more than or
        // equal to p
     
        if (x == 0)
          return 0; // In case x is divisible by p;
     
        while (y > 0)
        {
     
          // If y is odd, multiply x with result
          if ((y & 1) != 0)
            res = (res * x) % p;
     
          // y must be even now
          y = y >> 1; // y = y/2
          x = (x * x) % p;
        }
        return res;
      }
  
    
     // SIMPLE POWER FUNCTION=>
  
  
        static long power(long x, long y)
          {
              long res = 1;     // Initialize result
          
              while (y > 0)
              {
              
                  // If y is odd, multiply x with result
                  if ((y & 1) != 0)
                      res = res * x;
          
                  // y must be even now
                  y = y >> 1; // y = y/2
                  x = x * x;  // Change x to x^2
              }
              return res;
          }
  
  
  
  
  
  }