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.util.*;
import java.io.*;

public class Solution
{
    static class Reader {
		BufferedReader br;
		StringTokenizer st;
 
		public Reader() {
			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 class Ele implements Comparable<Ele> 
    {  
        public int x,y;  
        Ele(int x,int y) 
        {  
            this.x=x;this.y=y; 
        }  
        public int compareTo(Ele ob) {  
        if(ob.x!=x)return x-ob.x;
        return this.y-ob.y;  
        }   
    }
    int a[][];int b[][];
    
    long gcd(long a,long b)
    {
        long min=Math.min(a,b);
        long max=Math.max(a,b);
        while (max%min!=0)
        {
            a=max%min;
            max=min;min=a;
        }
        return min;
    }
	public static void main(String[] args) throws IOException 
	{
		Reader sc=new Reader();Solution G=new Solution();
		PrintWriter o = new PrintWriter(System.out);
		//int t=1;
		int t=sc.nextInt();
		
		int n,m,y,x0,y1,x1,x2,y2;int a[],b[],in[],in1[];long k,l;
		boolean b1[],b2;String s1[],s;
		//long l;long a[]; 
		ArrayList<ArrayList<Integer>> ll=new ArrayList<>();
		ArrayList<Integer> a1=new ArrayList<>();
		ArrayList<Integer> a2=new ArrayList<>();
		ArrayList<Integer> a3=new ArrayList<>();
		ArrayDeque<Integer> deq=new ArrayDeque<>();
		TreeSet<Integer> h0=new TreeSet<>();
		TreeSet<Integer> h1=new TreeSet<>();
		HashMap<Integer,Integer> h=new HashMap<>();
		try{
		while (t-->0)
		{
		    n=sc.nextInt();s=sc.next();a=new int[n];b1=new boolean[n];
		    for(int i=0;i<n-1;i++)
		    {
		        a1.add(s.charAt(i)-'0');
		    }
		    for (int i=n-1;i<2*n-1;i++)
		    {
		        
		        a1.add(s.charAt(i)-'0');x0=0;x1=0;
		        for (int j=0;j<n;j++)
		        {
		            if (!b1[j])
		            {
		                if (a1.get(j)==1)x1++;
		                else x0++;
		            }
		        }
		        if (x1>x0)
		        {
		            a2.add(1);
		            for (int j=0;j<n;j++)
    		        {
    		            if (!b1[j] && a1.get(j)==1)
    		            {
    		                b1[j]=true;
    		            }
    		        }
		        }
		        else
		        {
		            a2.add(0);
		            for (int j=0;j<n;j++)
    		        {
    		            if (!b1[j] && a1.get(j)==0)
    		            {
    		                b1[j]=true;
    		            }
    		        }
		        }
		        a1.remove(0);
		    }
		    //o.println(Arrays.toString(b));
		    for (int i:a2) o.print(i);
		    o.println();
		    //o.println(s);
		    //o.println();
		    //o.println();
		    //o.println(h);
		    //o.println(x2);
		    //o.println();
		    h0.clear();ll.clear();a1.clear();a2.clear();h1.clear();
		}
		}
		catch (Throwable e)
		{
		    e.printStackTrace();
		}
		//o.println("HI");
		
        o.flush();
        o.close();
	}
} 