human
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. 
Solve the task in PYTHON3.
assistant
#Here is Python code to print the corresponding binary string w of length n.

from sys import stdin, stdout
import math,sys,heapq
from itertools import permutations, combinations
from collections import defaultdict,deque,OrderedDict
from os import path
import bisect as bi
def yes():print('YES')
def no():print('NO')
if (path.exists('input.txt')): 
    #------------------Sublime--------------------------------------#
    sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
    def I():return (int(input()))
    def In():return(map(int,input().split()))
else:
    #------------------PYPY FAst I/o--------------------------------#
    def I():return (int(stdin.readline()))
    def In():return(map(int,stdin.readline().split()))
#sys.setrecursionlimit(1500)
def dict(a):
    d={} 
    for x in a:
        if d.get(x,-1)!=-1:
            d[x]+=1
        else:
            d[x]=1
    return d

def find_gt(a, x):
    'Find leftmost value greater than x'
    i = bi.bisect_right(a, x)
    if i != len(a):
        return i
    else:            
        return -1 

def main():
    try:
        n=I()
        l=list(input())
        for x in range(0,2*n,2):
            print(l[x],end="")
        print()
    except: 
        pass
        
M = 998244353
P = 1000000007
Inf=1000000000000000000
if __name__ == '__main__':
    for _ in range(I()):main()
    #for _ in range(1):main()