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.io.PrintWriter;
  import java.text.DecimalFormat;
  import java.util.*;
  
   
  public class Codeforces {
  	static int mod= 1000000007;
  	
  	public static void main(String[] args) throws Exception {
  		PrintWriter out=new PrintWriter(System.out);
  		FastScanner fs=new FastScanner();
  		int t=fs.nextInt();
  		
  		outer:while(t-->0) {
  			int s=fs.nextInt(), n=fs.nextInt();
  			int joy[][]=new int[n][s];
  			for(int i=0;i<s;i++) {
  				for(int j=0;j<n;j++)
  				joy[j][i]=fs.nextInt();
  			}
  			int ans=0;
  			int max[]=new int[n];
  			for(int i=0;i<n;i++) {
  				int cur=0;
  				for(int j=0;j<s;j++) {
  					cur=Math.max(cur,joy[i][j]);
  				}
  				max[i]=cur;
  			}
  			int l=1,r=1000000000;
  			while(l<r) {
  				int mid= (l+r+1)/2;
  				if(check(joy,mid,max)) {
  					l=mid;
  				}
  				else r=mid-1;
  			}
  			ans=l;
  			out.println(ans);
  		}
  		
  		out.close();
  		
  	}
  	static boolean check(int joy[][],int x,int []max) {
  		int n=joy.length, s=joy[0].length;
  		for(int i=0;i<n;i++) if(x>max[i]) return false;
  		for(int j=0;j<s;j++) {
  			int cnt=0;
  			for(int i=0;i<n;i++) {
  				if(joy[i][j]>=x) cnt++;
  			}
  			if(cnt>=2) return true;
  		}
  		return false;
  	}
  	static long pow(long a,long b) {
  		if(b<0) return 1;
  		long res=1;
  		while(b!=0) {
  			if((b&1)!=0) {
  				res*=a;
  				res%=mod;
  			}
  			a*=a;
  			a%=mod;
  			b=b>>1;
  		}
  		return res;
  	}
  	static int gcd(int  a,int  b) {
  		if(b==0) return a;
  		return gcd(b,a%b);
  	}
  	static long nck(int n,int k) {
  		if(k>n) return 0;
  		long res=1;
  		res*=fact(n);
  		res%=mod;
  		res*=modInv(fact(k));
  		res%=mod;
  		res*=modInv(fact(n-k)); 
  		res%=mod;
  		return res;
  	}
  	static long fact(long n) {
  //		return fact[(int)n];
  		long res=1;
  		for(int i=2;i<=n;i++) {
  			res*=i;
  			res%=mod;
  		}
  		return res;
  	}
  	
  	static long modInv(long n) {
  		return pow(n,mod-2);
  	}
  	
  	static void sort(int[] a) {
  		//suffle
  		int n=a.length;
  		Random r=new Random();
  		for (int i=0; i<a.length; i++) {
  			int oi=r.nextInt(n);
  			int temp=a[i];
  			a[i]=a[oi];
  			a[oi]=temp;
  		}
  		
  		//then sort
  		Arrays.sort(a);
  	}
  	
  	// Use this to input code since it is faster than a Scanner
  	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());
  		}
  		long[] lreadArray(int n) {
  			long a[]=new long[n];
  			for(int i=0;i<n;i++) a[i]=nextLong();
  			return a;
  		}
  		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());
  		}
  	}
   
  }