human
Write me a Python code to print the corresponding binary string w of length n given the following:
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. 
Using python can you solve the prior task?
assistant
#Here is Python code to print the corresponding binary string w of length n.

from __future__ import division, print_function

def main():
    # Template 1.0
    import sys, re, math
    from collections import deque, defaultdict, Counter, OrderedDict
    from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, floor
    from heapq import heappush, heappop, heapify, nlargest, nsmallest
    def STR(): return list(input())

    def INT(): return int(input())

    def MAP(): return map(int, input().split())

    def LIST(): return list(map(int, input().split()))

    def list2d(a, b, c): return [[c] * b for i in range(a)]

    def sortListWithIndex(listOfTuples, idx):   return (sorted(listOfTuples, key=lambda x: x[idx]))

    def sortDictWithVal(passedDic):
        temp = sorted(passedDic.items(), key=lambda kv: (kv[1], kv[0]))
        toret = {}
        for tup in temp:
            toret[tup[0]] = tup[1]
        return toret

    def sortDictWithKey(passedDic):
        return dict(OrderedDict(sorted(passedDic.items())))

    INF = float('inf')
    mod = 10 ** 9 + 7

    t = INT()

    while(t!=0):
        n = INT()
        s= input()
        ans = ""
        for i in range(0, 2*n-1, 2):
            ans+=s[i]
        print(ans)
        t-=1



######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip

    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192

class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def...