A binary string is a string where each character is either 0 or 1. Two binary strings a and b of equal length are similar, if they have the same character in some position (there exists an integer i such that a_i = b_i). For example:

  * 10010 and 01111 are similar (they have the same character in position 4); 
  * 10010 and 11111 are similar; 
  * 111 and 111 are similar; 
  * 0110 and 1001 are not similar. 



You are given an integer n and a binary string s consisting of 2n-1 characters. Let's denote s[l..r] as the contiguous substring of s starting with l-th character and ending with r-th character (in other words, s[l..r] = s_l s_{l + 1} s_{l + 2} ... s_r).

You have to construct a binary string w of length n which is similar to all of the following strings: s[1..n], s[2..n+1], s[3..n+2], ..., s[n..2n-1].

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first line of each test case contains a single integer n (1 ≤ n ≤ 50).

The second line of each test case contains the binary string s of length 2n - 1. Each character s_i is either 0 or 1.

Output

For each test case, print the corresponding binary string w of length n. If there are multiple such strings — print any of them. It can be shown that at least one string w meeting the constraints always exists.

Example

Input


4
1
1
3
00000
4
1110000
2
101


Output


1
000
1010
00

Note

The explanation of the sample case (equal characters in equal positions are bold):

The first test case: 

  * 1 is similar to s[1..1] = 1. 



The second test case: 

  * 000 is similar to s[1..3] = 000; 
  * 000 is similar to s[2..4] = 000; 
  * 000 is similar to s[3..5] = 000. 



The third test case: 

  * 1010 is similar to s[1..4] = 1110; 
  * 1010 is similar to s[2..5] = 1100; 
  * 1010 is similar to s[3..6] = 1000; 
  * 1010 is similar to s[4..7] = 0000. 



The fourth test case: 

  * 00 is similar to s[1..2] = 10; 
  * 00 is similar to s[2..3] = 01. 
import java.io.*;
import java.util.*;

public class Main{
    public static void main(String[] args) {
        FastReader fs=new FastReader();
        PrintWriter out=new PrintWriter(System.out);
        int t=1;
        t=fs.nextInt();
        for(int i=1;i<=t;i++) {
        	Main m=new Main();
            m.solve(i,out,fs);
        }
        out.close();
    } 
    ArrayList<ArrayList<Integer>> adj;
    int tin[],tout[];
    int timer;
    void dfs(int u,int p) {
    	tin[u]=++timer;
    	for(int v:adj.get(u)) {
    		if(v==p)continue;
    		dfs(v,u);
    	}
    	tout[u]=timer;
    }
    boolean isAncestor(int u,int v) {
    	return tin[u]<=tin[v]&&tout[u]>=tout[v];
    }
    public void solve(int test,PrintWriter out,FastReader fs){
        int n=fs.nextInt();
        char c[]=fs.nextCharArray();
        for(int i=0;i<2*n-1;i+=2){
            out.print(c[i]);
        }
        out.println();
    }
}
class SegmentTree{
	long t[];
	SegmentTree(int n){
		t=new long[4*n];
	}
	void update(int v,int tl,int tr,int pos,long delta) {
		if(tl==tr) {
			t[v]+=delta;
		}else {
			int tm=(tl+tr)/2;
			if(pos<=tm)
				update(2*v,tl,tm,pos,delta);
			else
				update(2*v+1,tm+1,tr,pos,delta);
			t[v]=t[2*v]+t[2*v+1];
		}
	}
	long sum(int v,int l,int r,int tl,int tr) {
		if(l>r)return 0;
		if(l==tl&&r==tr) {
			return t[v];
		}else {
			int tm=(tl+tr)/2;
			return sum(2*v,l,Integer.min(r,tm),tl,tm)+sum(2*v+1,Integer.max(l, tm+1),r,tm+1,tr);
		}
	}
}
class FastReader{
    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 nextLong(){
        return Long.parseLong(next());
    }
    int[] nextIntArray(int n){
        int a[]=new int[n];
        for(int i=0;i<n;i++)a[i]=nextInt();
        return a;
    } 
    long[] nextLongArray(int n){
        long a[]=new long[n];
        for(int i=0;i<n;i++)a[i]=nextInt();
        return a;
    }
    char[] nextCharArray(){
        char a[]=next().toCharArray();
        return a;
    }
}