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 math 
from collections import deque
import sys


sys.setrecursionlimit(10**4) 

def Divisors(n) : 
	
	l=[]
	i = 2
	while i <= math.sqrt(n): 
		
		if (n % i == 0) : 
			
			
			if (n // i == i) : 
				l.append(i)
			else : 
				l.append(i)
				l.append(n//i)
		i = i + 1
	return l

def SieveOfEratosthenes(n): 
	
	l=[]
	
	prime = [True for i in range(n+1)] 
	p = 2
	while (p * p <= n): 
		
		if (prime[p] == True): 
			
			for i in range(p * p, n+1, p): 
				prime[i] = False
		p += 1
	
	for p in range(2, n+1): 
		if prime[p]: 
			l.append(p)
	
	return l		

def primeFactors(n): 
	
	l=[]
	
	while n % 2 == 0: 
		l.append(2) 
		n = n / 2
		
	for i in range(3,int(math.sqrt(n))+1,2): 
		
		while n % i== 0: 
			l.append(i) 
			n = n / i 
			
	if n > 2: 
		l.append(n)
	
	return(l)	
		

def Factors(n) : 
	
	
	result = []
	
	for i in range(2,(int)(math.sqrt(n))+1) : 

		if (n % i == 0) : 
		    
			if (i == (n/i)) : 
				result.append(i)
			else : 
			    result.append(i)
			    result.append(n//i)
		
	result.append(1)
	
	return result

def maxSubArraySum(a): 
	
	max_so_far = 0
	max_ending_here = 0
	size=len(a)
	
	for i in range(0, size): 
		max_ending_here = max_ending_here + a[i] 
		if (max_so_far < abs(max_ending_here)): 
			max_so_far = max_ending_here 

	return max_so_far 

def longestsubarray(arr, n, k): 
    current_count = 0
      
    # this will contain length of  
    # longest subarray found 
    max_count = 0
  
    for i in range(0, n, 1): 
        if (arr[i] % k != 0): 
            current_count += 1
        else: 
            current_count = 0
        max_count = max(current_count,  
                            max_count) 
      
    return max_count 
    
#print(SieveOfEratosthenes(100))
#print(Divisors(100))
#print(primeFactors(100))
#print(Factors(100))
#print(maxSubArraySum(a))
 

def main():
    
    n=int(input())
    s=input()
    s1=""
    for i in range(0,len(s),2):
        s1+=s[i]
    print(s1)
    
    
    
t=int(input())
for i in range(0,t):
    main()