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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.*;
  
  public class CF1 {
      public static void main(String[] args) {
          FastScanner sc=new FastScanner();
          int T=sc.nextInt();
  //        int T=1;
          for (int tt=0; tt<T; tt++){
             int m = sc.nextInt();
             int n = sc.nextInt();
             int maxes[]= new int [n];
             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();
                     maxes[j]=Math.max(maxes[j],arr[i][j]);
                 }
             }
             for (int i=0; i<m; i++){
                 Arrays.sort(arr[i]);
             }
             int min=maxes[0];
             for (int i=0; i<n; i++){
                 min=Math.min(maxes[i],min);
             }
             int ans =0;
             for (int i=0; i<m; i++){
                 ans=Math.max(ans, Math.min(min, arr[i][n-2]));
             }
              System.out.println(ans);
          }
      }
      static long mod =998244353L;
      static long power (long a, long b){
          long res=1;
          while (b>0){
              if ((b&1)== 1){
                  res= (res * a % mod)%mod;
              }
              a=(a%mod * a%mod)%mod;
              b=b>>1;
          }
          return res;
      }
  
      boolean[] sieveOfEratosthenes(int n)
      {
  
          boolean prime[] = new boolean[n+1];
          for(int i=0;i<=n;i++)
              prime[i] = true;
  
          for(int p = 2; p*p <=n; p++)
          {
              if(prime[p] == true)
              {
                  for(int i = p*p; i <= n; i += p)
                      prime[i] = false;
              }
          }
          return prime;
      }
      static void sort(int[] a) {
          ArrayList<Integer> l=new ArrayList<>();
          for (int i:a) l.add(i);
          Collections.sort(l);
          for (int i=0; i<a.length; i++) a[i]=l.get(i);
      }
      static void sortLong(long[] a) {
          ArrayList<Long> l=new ArrayList<>();
          for (long i:a) l.add(i);
          Collections.sort(l);
          for (int i=0; i<a.length; i++) a[i]=l.get(i);
      }
      static long gcd (long n, long m){
          if (m==0) return n;
          else return gcd(m, n%m);
      }
  
      static class Pair implements Comparable<Pair>{
          int x,y;
          public Pair(int x, int y){
              this.x = x;
              this.y = y;
          }
          public int compareTo(Pair o){
              return this.x-o.x;
          }
      }
      static class FastScanner {
          BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
          StringTokenizer st=new StringTokenizer("");
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st=new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
          int[] readArray(int n) {
              int[] a=new int[n];
              for (int i=0; i<n; i++) a[i]=nextInt();
              return a;
          }
          long nextLong() {
              return Long.parseLong(next());
          }
      }
  }