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
import math
import itertools
import functools
import collections
import operator
import fileinput
import copy
 
ORDA = 97  # a
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return [int(i) for i in input().split()]
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def revn(n): return str(n)[::-1]
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
def sieve(n):
    if n < 2: return list()
    prime = [True for _ in range(n + 1)]
    p = 3
    while p * p <= n:
        if prime[p]:
            for i in range(p * 2, n + 1, p):
                prime[i] = False
        p += 2
    r = [2]
    for p in range(3, n + 1, 2):
        if prime[p]:
            r.append(p)
    return r
def divs(n, start=2):
    r = []
    for i in range(start, int(math.sqrt(n) + 1)):
        if (n % i == 0):
            if (n / i == i):
                r.append(i)
            else:
                r.extend([i, n // i])
    return r
def divn(n, primes):
    divs_number = 1
    for i in primes:
        if n == 1:
            return divs_number
        t = 1
        while n % i == 0:
            t += 1
            n //= i
        divs_number *= t
def prime(n):
    if n == 2: return True
    if n % 2 == 0 or n <= 1: return False
    sqr = int(math.sqrt(n)) + 1
    for d in range(3, sqr, 2):
        if n % d == 0: return False
    return True
def convn(number, base):
    new_number = 0
    while number > 0:
        new_number += number % base
        number //= base
    return new_number
def cdiv(n, k): return n // k + (n % k != 0)
def ispal(s): # Palindrome
    for i in range(len(s) // 2 + 1):
        if s[i] != s[-i - 1]:
            return False
    return True
  
# a = [1,2,3,4,5] -----> print(*a) ----> list print krne ka new way

for i in range(ii()):
	n = ii()
	s = input()
	print(s[n-1]*n)