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,bisect

from sys  import stdin,stdout

from bisect import bisect_left,bisect_right,bisect,insort,insort_left,insort_right

from math import gcd,ceil,floor,sqrt

from collections import Counter,defaultdict,deque,OrderedDict

from queue import Queue,PriorityQueue

from string import ascii_lowercase


sys.setrecursionlimit(10**6)
INF = float('inf')
MOD = 998244353
mod = 10**9+7

def isPrime(n):
    if (n <= 1) :return False
    if (n <= 3) :return True
    if (n%2 == 0 or n%3 == 0):return False
    for i in range(5,ceil(sqrt(n))+1,6):
        if (n%i==0 or n%(i+2)==0):
            return False
    return True

def st():
    return list(stdin.readline().strip())

def inp():
    return int(stdin.readline())

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

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

def pr(n):
    stdout.write(str(n)+"\n")
    
def DFS(dictionary,vertex,visited):
    visited[vertex]=True
    stack=[vertex]
    print(vertex)
    while stack:
        a=stack.pop()
        for i in dictionary[a]:
            if not visited[i]:
                print(i)
                visited[i]=True
                stack.append(i)


def BFS(dictionary, vertex,visited):
    visited[vertex]=True
    q=deque()
    q.append(vertex)
    while q:
        a=q.popleft()
        for i in dictionary[a]:
            if not visited[i]:
                visited[i]=True
                q.append(i)
                print(i)

                


def soe(limit):
    l=[1]*(limit+1)
    l[0]=0
    l[1]=0
    prime=[]
    for i in range(2,limit+1):
        if l[i]:
            for j in range(i*i,limit+1,i):
                l[j]=0
    
    for i in range(2,limit+1):
        if l[i]:
            prime.append(i)
    return prime

def segsoe(low,high):
    limit=int(high**0.5)+1
    prime=soe(limit)
    n=high-low+1
    l=[0]*(n+1)
    for i in range(len(prime)):
        lowlimit=(low//prime[i])*prime[i]
        if lowlimit<low:
            lowlimit+=prime[i]
        if lowlimit==prime[i]:
            lowlimit+=prime[i]
        for j in range(lowlimit,high+1,prime[i]):
            l[j-low]=1
    for i in range(low,high+1):
        if not l[i-low]:
            if i!=1:
                print(i)
                
def gcd(a,b):
    while b:
        a=a%b
        b,a=a,b
    return a

def power(a,n):
    r=1
    while n:
        if n&1:
            r=(r*a)
        a*=a
        n=n>>1
    return r
    
def solve():
    n=inp()
    s=st()
    for i in range(n):
        a=s[i:i+n]
        z=a.count('1')
        y=a.count('0')
        if z==0 :
            pr('0'*n)
            return
        if y==0:
            pr('1'*n)
            return
    pr('0'*n)
    
    



for _ in range(inp()):
    solve()
