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. 
from collections import defaultdict as dc
import sys
from collections import Counter as cnt
from collections import deque as dq
from copy import copy,deepcopy
import itertools 
import math
mod=10**9 +7
# How to sort a list using two compare :
# instead of defining a function define class as a key
# class key(str):
#     def __lt__(x,y):
#         return int(x+y)>int(y+x)
def sinp():
    p=str(sys.stdin.readline())
    return p

def seive(n):
    p=[0]*(n+1)
    for i in range(2,n):           
        for j in range(2*i,n+1,i):
                p[i]=1
    return p

def inp(): 
    p=int(sys.stdin.readline())
    return p

def line():
    p=list(map(int,sys.stdin.readline().split()))
    return p

def read_mat():
    n=inp()
    a=[]
    for i in range(n):
        a.append(line())
    return a

def ind(a):
    return ord(a)-ord('A')

def digit(n):
    p=[int(i) for i in str(n)]
    return p

def check(p,n):
    l=str(p)
    s=''
    for i in range(n):
        s=s+bin(int(l[i]))[2:]
    s=s[:-n]
    if s=='':
        return 0
    return int(s,2)

def createcomb(a,n,i):
    if i==n:
        ans.append(a)
    else:
        for z in range(n):
            a.append(a[z])
            createcomb(a,n,z)  
            a.pop()          

def solve(a,n,a1,b1,c1):
    l=[]
    r=[]
    x=0
    for i in range(n):
        if a[i]>=x:
            l.append(i)
            x=a[i]
    x=0
    for i in range(n-1,-1,-1):
        if a[i]>=x:
            r.append(i)
            x=a[i]
    z=cnt(r)
    c=0
    p=len(l)
    q=len(r)
    for i in l:
        if z[i]:
            c+=1
    if p==a1 and q==b1 and c1==c:
        return 1
    return 0

def xor(a,b):
    s=''
    for i in range(len(a)):
        if a[i]=='1' and b[i]=='0':
            s+='1'
        elif b[i]=='1' and a[i]=='0':
            s+='1'
        else:
            s+='0'
    return s
for t in range(inp()):
    n=inp()
    a=str(input())
    s=''
    for i in range(len(a)-n+1):
        s+=a[i:i+n][i]
    print(s)