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 Main
  {
      static PrintWriter out = new PrintWriter((System.out));
      static Kioken sc = new Kioken();
  
      public static void main(String[] args)
      {
        int t = 1;
        t = sc.nextInt();
        while (t-- > 0)
        {
            solve();
        }
        out.close();
      }
  
      static boolean bs(int a, int[][] arr, int n, int m){
        
        int cnt = 0;
        int min_travel = Math.min(n-1, m);
        HashSet<Integer> shop = new HashSet<>();
  
        for(int j = 0; j < n; j++){
          boolean flag = true;
          HashSet<Integer> set = new HashSet<>();
          for(int i = 0; i < m; i++){
            if(arr[i][j]>= a){
              set.add(i);
            }
          }
  
          if(set.isEmpty()) return false;
  
          // check if there is any shop common from previous
          for(int x: set){
            if(shop.contains(x)){
              flag = false;
              break;
            }
          }
  
          if(flag){
            // no common shop available till now
            shop.addAll(set);
            cnt++;
            // return false;
          }
        }
  
        if(cnt <= min_travel) return true;
        return false;
      }
      public static void solve()
      {
        int m = sc.nextInt();
        int n = sc.nextInt();
        
        int r = Integer.MIN_VALUE;
        int l = Integer.MAX_VALUE;
        int[][] arr = new int[m][n];
        for(int i = 0; i < m; i++){
          for(int j = 0; j < n; j++){
            arr[i][j] = sc.nextInt();
            r = Math.max(r, arr[i][j]);
            l = Math.min(l, arr[i][j]);
          }
        }
  
   
        // out.println(" l " + l  + " " + r);
        int ans = 0;
        while(l <= r){
          int mid = (l+r)/2;
          // out.println(" mid " + mid);
          if(bs(mid, arr, n, m)){
            ans = mid;
            l = mid+1;
          }else{
            r = mid-1;
          }
        }
        out.println(ans);
      }
  
      public static long leftShift(long a){
          return (long)Math.pow(2, a);
      }
  
      
      public static void reverse(int[] arr) {
      	Arrays.sort(arr);
      	int n = arr.length;
      	for(int i = 0 ; i < arr.length; i++) {
      		int temp = arr[i];
      		arr[i] = arr[n - 1 - i];
      		arr[n - 1 - i] = temp;
      	}
      	return ;
      }
      
      public static int lower_bound(ArrayList<Integer> ar, int k)
      {
          int s = 0, e = ar.size();
          while (s != e)
          {
              int mid = s + e >> 1;
              if (ar.get(mid) <= k)
              {
                  s = mid + 1;
              }
              else
              {
                  e = mid;
              }
          }
          return Math.abs(s) - 1;
      }
  
      public static int upper_bound(ArrayList<Integer> ar, int k)
      {
          int s = 0;
          int e = ar.size();
          while (s != e)
          {
              int mid = s + e >> 1;
              if (ar.get(mid) < k)
              {
                  s = mid + 1;
              }
              else
              {
                  e = mid;
              }
          }
          if (s == ar.size())
          {
              return -1;
          }
          return s;
      }
  
      static class Kioken
      {
          // FileInputStream br = new FileInputStream("input.txt");
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st = new StringTokenizer("");
  
          public String next()
          {
              while (!st.hasMoreTokens())
              {
                  try
                  {
                      st = new StringTokenizer(br.readLine());
                  }
                  catch (Exception e)
                  {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          public int nextInt()
          {
              return Integer.parseInt(next());
          }
  
          public long nextLong()
          {
              return Long.parseLong(next());
          }
  
          public double nextDouble()
          {
              return Double.parseDouble(next());
          }
  
          public String nextLine()
          {
              try
              {
                  return br.readLine();
              }
              catch (Exception e)
              {
                  e.printStackTrace();
              }
              return null;
          }
  
          public boolean hasNext()
          {
              String next = null;
              try
              {
                  next = br.readLine();
              }
              catch (Exception e)
              {
              }
              if (next == null || next.length() == 0)
              {
                  return false;
              }
              st = new StringTokenizer(next);
              return true;
          }
      }
  }