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. 

Write me a Python code to print the corresponding binary string w of length n?
assistant
#Here is Python code to print the corresponding binary string w of length n.

# It's never too late to start!
from bisect import bisect_left
import os
import sys
from io import BytesIO, IOBase
from collections import Counter, defaultdict
from collections import deque
from functools import cmp_to_key
import math
# import heapq

def sin():
    return input()
def ain():
    return list(map(int, sin().split()))
def sain():
    return input().split()
def iin():
    return int(sin())
MAX = float('inf')
MIN = float('-inf')
MOD = 1000000007
def sieve(n): 
    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
    s = set()
    for p in range(2, n+1): 
        if prime[p]: 
            s.add(p)
    return s
def readTree(n, m):
    adj = [deque([]) for _ in range(n+1)]
    for _ in range(m):
        u,v = ain()
        adj[u].append(v)
        adj[v].append(u)
    return adj

def solve(c1,c2,maxi1,maxi2):
    for i in range(1,maxi1+1):
        if i not in c1:
            return False
    for i in range(1,maxi2+1):
        if i not in c2:
            return False
    return True
# Stay hungry, stay foolish!

def main():
    for _ in range(iin()):
        n = iin()
        s = sin()
        l = [0]*n
        for i in range(0,n):
            x = s[i:i+n]
            # print(x)
            for i in range(n):
                l[i] += int(x[i])
        for i in range(n):
            if l[i] == n:
                l[i] = str(1)
            else:
                l[i] = str(0)
        print("".join(l))

        




# Fast IO Template starts

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
        ...