On the competitive programming platform CodeCook, every person has a rating graph described by an array of integers a of length n. You are now updating the infrastructure, so you've created a program to compress these graphs.

The program works as follows. Given an integer parameter k, the program takes the minimum of each contiguous subarray of length k in a.

More formally, for an array a of length n and an integer k, define the k-compression array of a as an array b of length n-k+1, such that $$$b_j =min_{j≤ i≤ j+k-1}a_i$$$

For example, the 3-compression array of [1, 3, 4, 5, 2] is [min\{1, 3, 4\}, min\{3, 4, 5\}, min\{4, 5, 2\}]=[1, 3, 2].

A permutation of length m is an array consisting of m distinct integers from 1 to m in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (m=3 but there is 4 in the array).

A k-compression array will make CodeCook users happy if it will be a permutation. Given an array a, determine for all 1≤ k≤ n if CodeCook users will be happy after a k-compression of this array or not.

Input

The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.

The first line of the description of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the array.

The second line of the description of each test case contains n integers a_1,…,a_n (1≤ a_i≤ n) — the elements of the array.

It is guaranteed, that the sum of n for all test cases does not exceed 3⋅ 10^5.

Output

For each test case, print a binary string of length n. 

The k-th character of the string should be 1 if CodeCook users will be happy after a k-compression of the array a, and 0 otherwise. 

Example

Input


5
5
1 5 3 4 2
4
1 3 2 1
5
1 3 3 3 2
10
1 2 3 4 5 6 7 8 9 10
3
3 3 2


Output


10111
0001
00111
1111111111
000

Note

In the first test case, a=[1, 5, 3, 4, 2].

  * The 1-compression of a is [1, 5, 3, 4, 2] and it is a permutation. 
  * The 2-compression of a is [1, 3, 3, 2] and it is not a permutation, since 3 appears twice. 
  * The 3-compression of a is [1, 3, 2] and it is a permutation. 
  * The 4-compression of a is [1, 2] and it is a permutation. 
  * The 5-compression of a is [1] and it is a permutation. 
import sys
import math
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log2, ceil
from collections import defaultdict as dd
from bisect import bisect_left as bl, bisect_right as br
from bisect import insort
from collections import Counter
from collections import deque
from heapq import heappush,heappop,heapify
from itertools import permutations,combinations
mod = int(1e9)+7
mod = 998244353
 
 
ip = lambda : int(stdin.readline())
inp = lambda: map(int,stdin.readline().split())
ips = lambda: stdin.readline().rstrip()
out = lambda x : stdout.write(str(x)+"\n")


 
def check(ch):
    ch.sort()
    ct = 1
    flag = True
    for i in range(len(ch)):
        if i+1 == ch[i]:
            pass
        else:
            flag = False
            break
    return flag

t = ip()
for _ in range(t):
    n = ip()
    arr = list(inp())
    tree = sorted(list(arr))
    tree = deque(tree)
    ans = [None]*n
    st = 0
    end = n-1
    i = 0
    if check(list(arr)):
        ans[n-1] = '1'
    while i<n-1:
        if arr[st] == i+1:
            st += 1
            i += 1
            tree.popleft()
        else:
            if arr[end] == i+1:
                end -= 1
                i += 1
                tree.popleft()
            else:
                break
        val = tree[0]
        if val == i+1:
            ans[i] = '1'
        else:
            break
    ans = ans[::-1]
    if min(arr) == 1:
        ans[n-1] = '1'
    for i in range(n):
        if ans[i] == None:
            ans[i] = '0'
    
    ans = ''.join(ans)
    out(ans)
        
        
    
        
                    
                    
                    
        
                
                
                    
                    
                
            
                
            

    
        
       
        


    
    
    
        
            
        
        
        
        
    
        
            
    
        
            
        
    
            
        
        
    

    
    
        
            
        
            
        
        
            
            
        
    
        
                    
                 
    
    
    
    
    
    
    
            
                
                
    
        
    
            
    
                    
                    
            
    
            
                
                
            
    
        
        
                    
        
        
            
            
            
        

        
    
        
    
            
    
    
        
                
        
        
            
                        
                
                    
                
                    
                
        
    
        
    
    
    
                        
            
            
    
            
            
        
            
                    
            
                    
                    
                    
                    
                    
                    
                    
    
        
    
    
    
    
    
            
        
                
        
            
        
    
    
    
            
            
        
    
            
            
        
                
                
        
    
        
        
    
    
            

        
        
        

            
            
        
            
        
        
        
            
        
    
            
        
                
        
        
            
            
    
                
        
                
            
            
        
    
        
    
                
        
    
    
    
    
    
    
            

    
            
            
            
        
    

            
        
