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
from functools import reduce
from collections import Counter
import time
import datetime
from math import sqrt,gcd

# def time_t():
#     print("Current date and time: " , datetime.datetime.now())
#     print("Current year: ", datetime.date.today().strftime("%Y"))
#     print("Month of year: ", datetime.date.today().strftime("%B"))
#     print("Week number of the year: ", datetime.date.today().strftime("%W"))
#     print("Weekday of the week: ", datetime.date.today().strftime("%w"))
#     print("Day of year: ", datetime.date.today().strftime("%j"))
#     print("Day of the month : ", datetime.date.today().strftime("%d"))
#     print("Day of week: ", datetime.date.today().strftime("%A"))

def ip(): return int(sys.stdin.readline())

def sip(): return sys.stdin.readline()

def mip(): return map(int,sys.stdin.readline().split())

def mips(): return map(str,sys.stdin.readline().split())

def lip(): return list(map(int,sys.stdin.readline().split()))

def matip(n,m):
    lst=[]
    for i in range(n):
        arr = lip()
        lst.append(arr)
    return lst

def factors(n): # find the factors of a number
    return list(set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))

def minJumps(arr, n): #to reach from 0 to n-1 in the array in minimum steps
    jumps = [0 for i in range(n)]
    if (n == 0) or (arr[0] == 0):
        return float('inf')
    jumps[0] = 0
    for i in range(1, n):
        jumps[i] = float('inf')
        for j in range(i):
            if (i <= j + arr[j]) and (jumps[j] != float('inf')):
                jumps[i] = min(jumps[i], jumps[j] + 1)
                break
    return jumps[n-1]

def dic(arr): # converting list into dict of count
    return Counter(arr)

def check_prime(n):
    if n<2:
        return False
    for i in range(2,int(n**(0.5))+1,2):
        if n%i==0:
            return False
    return True 

# --------------------------------------------------------- #
# sys.stdin = open('input.txt','r')
# sys.stdout = open('output.txt','w')
# --------------------------------------------------------- #

t = ip()
while t:
    t-=1
    n = ip()
    s = sip()
    if n==1:
        print(s)
    elif n==2:
        print(s[1]+s[1])
    else:
        k = s[((2*n)-1)//2]
        ans = k*n
        print(ans)
