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: 
  
  // D
  import java.util.*;
  import java.io.*;
  public class Solution {
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
          public FastReader() {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
          int nextInt() {
              return Integer.parseInt(next());
          }
          long nextLong() {
              return Long.parseLong(next());
          }
          double nextDouble() {
              return Double.parseDouble(next());
          }
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
      static void sort(int a[]){ // int -> long
          ArrayList<Integer> arr=new ArrayList<>(); // Integer -> Long
          for(int i=0;i<a.length;i++)
          arr.add(a[i]);
          Collections.sort(arr);
          for(int i=0;i<a.length;i++)
          a[i]=arr.get(i);
          
      }
      private static long gcd(long a, long b){
          if(b==0)return a;
          return gcd(b,a%b);
      }
      private static long pow(long x,long y){
          if(y==0)return 1;
          long temp = pow(x, y/2);
          if(y%2==1){
              return x*temp*temp;
          }
          else{
              return temp*temp;
          }
      }
      static int log(long n){
          int res = 0;
          while(n>0){
              res++;
              n/=2;
          }
          return res;
      }
  
      static int mod = (int)1e9+7;
      static PrintWriter out;
      static FastReader sc ;
      public static void main(String[] args) throws IOException {
          sc = new FastReader();
          out = new PrintWriter(System.out);
          // primes();
          // ================================ //
          int test = sc.nextInt();
          while (test-- > 0) {
              int n = sc.nextInt();
              int m = sc.nextInt();
              int[][] mat = new int[n][m];
              for(int i=0;i<n;i++)
                  for(int j=0;j<m;j++)
                      mat[i][j] = sc.nextInt();
              solver(mat, n, m);
          }
          // ================================ //
          // int n = sc.nextInt();
          // solver();
          // ================================ //
          out.flush();
      }
  
      public static void solver(int[][]mat, int n, int m) {
          int[]best = new int[m];
          for(int i=0;i<n;i++){
              for(int j=0;j<m;j++){
                  best[j] = Math.max(best[j], mat[i][j]);
              }
          }
          int res = 0;
          for(int i=0;i<n;i++){
              int[][]temp = new int[m][2];
              for(int j=0;j<m;j++){
                  temp[j][1] = j;
                  temp[j][0] = mat[i][j];
              }
              Arrays.sort(temp, (a,b)->b[0]==a[0]?best[b[1]]-best[a[1]]:b[0]-a[0]);
              int ans = temp[1][0]; 
              for(int j=0;j<m;j++){
                  if(j==temp[0][1] || j==temp[1][1])continue; 
                  ans = Math.min(ans, best[j]);
              }
              // System.out.println(i+"__"+    ans);
              res = Math.max(res, ans);
          }
          out.println(res);
      }
  }
  
  
  