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 sys


#def solve(a,b,c,d):
def solve(S,n):
    '''ones = int(S[:n],2)
    zeroes = ~ones
    #print('cum:',cum)
    for i in range(1,n):
        si = int(S[i:(i+n)],2)
        #print('  ',si, S[i:(i+n)])
        ones &= si
        zeroes &= ~si
    print(S,n,ones,zeroes)
    return ~zeroes | ones



    cum = [set([S[i]]) for i in range(n)]
    while True:
        change = False
        for i in range(1,n):
            for j,c in enumerate(S[i:(i+n)]):
                if not c in cum[j]:
                    change = True
                    cum[j].add(c)
    print(cum)
    #return ''.join([list(a)[-1] for a in cum])
    for i in range(n):
        for j,c in enumerate(S[i:(i+n)]):
            cum[j].add(c)

    #return ''.join([list(a)[-1] for a in cum])



    start = ['0']*n
    for i in range(n):
        match = False
        for j,c in enumerate(S[i:(i+n)]):
            if start[j]==c:
                match = True
        if match:
            continue
        else:
            for j,c in enumerate(S[i:(i+n)]):
                if not start[j]==c:
                    start[j] = '1'
                    break
    '''
    C = S[n-1]*n
    return ''.join(C)

            
            
    


if __name__ == '__main__':
    IN = [x.strip() for x in sys.stdin.readlines()]

    T = int(IN[0])
    cur = 0
    for ti in range(T):
        #a = int(IN[cur])
        #a = [int(x) for x in IN[cur].split(' ')]
        n = int(IN[cur+1])
        S = IN[cur+2]#[int(x) for x in IN[cur+1].split(' ')]
        res = solve(S,n)
        print(res)
        cur += 2

