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: /*
   * Everything is Hard 
   * Before Easy 
   * Jai Mata Dii 
   */ 
   
  import java.util.*;
  import java.io.*; 
    
  public class Main {
  	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 long mod = (long)(1e9+7); 
  //	 static long mod = 998244353; 
  //	 static Scanner sc = new Scanner(System.in); 
  	static FastReader sc = new FastReader(); 
  	static PrintWriter out = new PrintWriter(System.out);
  	public static void main (String[] args) {
  		int ttt = 1;
  		ttt = sc.nextInt();
  		z :for(int tc=1;tc<=ttt;tc++){
  			int n = sc.nextInt();
  			int m = sc.nextInt();
  			long a[][] = new long[n][m];
  			for(int i=0;i<n;i++) {
  				for(int j=0;j<m;j++) {
  					a[i][j] = sc.nextLong();
  //					sort(a[i]);
  				}
  			}
  			
  			int f1 = 0, f2 = 0;
  			long min = 0;
  			for(int i=0;i<n;i++) {
  				long m1=0,m2=0; int k1=0,k2=0;
  				for(int j=0;j<m;j++) {
  					if(a[i][j]>m1) {
  						m2 = m1;
  						m1 = a[i][j];
  						k2=k1;
  						k1=j;
  					}
  					else if(a[i][j]>m2) {
  						m2 = a[i][j];
  						k2=j;
  					}
  				}
  				if(Math.min(m1, m2)>min) {
  					min = Math.min(m2, m1);
  					f1 = k1; f2 = k2;
  				}
  			}
  			long max[] = new long[m];
  			for(int i=0;i<n;i++) {
  				for(int j=0;j<m;j++) {
  					if(j==f1 || j == f2) continue;
  					max[j] = Math.max(max[j], a[i][j]);
  				}
  			}
  			for(int j=0;j<m;j++) {
  				if(j==f1 || j == f2) continue;
  				min = Math.min(min, max[j]);
  			}
  			out.write(min+"\n");
  		}
  		out.close();
  	}
  	static long pow(long a, long b){long ret = 1;while(b>0){if(b%2 == 0){a = (a*a)%mod;b /= 2;}else{ret = (ret*a)%mod;b--;}}return ret%mod;}
  	static long gcd(long a,long b){if(b==0) return  a; return gcd(b,a%b); } 
  	private static void sort(int[] a) {List<Integer> k = new ArrayList<>();for(int val : a) k.add(val);Collections.sort(k);for(int i=0;i<a.length;i++) a[i] = k.get(i);} 
  	private static void ini(List<Integer>[] tre2){for(int i=0;i<tre2.length;i++){tre2[i] = new ArrayList<>();}} 
  	private static void init(List<int[]>[] tre2){for(int i=0;i<tre2.length;i++){tre2[i] = new ArrayList<>();}} 
  	private static void sort(long[] a) {List<Long> k = new ArrayList<>();for(long val : a) k.add(val);Collections.sort(k);for(int i=0;i<a.length;i++) a[i] = k.get(i);} 
  }