You are given a permutation p=[p_1, p_2, …, p_n] of integers from 1 to n. Let's call the number m (1 ≤ m ≤ n) beautiful, if there exists two indices l, r (1 ≤ l ≤ r ≤ n), such that the numbers [p_l, p_{l+1}, …, p_r] is a permutation of numbers 1, 2, …, m.

For example, let p = [4, 5, 1, 3, 2, 6]. In this case, the numbers 1, 3, 5, 6 are beautiful and 2, 4 are not. It is because:

  * if l = 3 and r = 3 we will have a permutation [1] for m = 1; 
  * if l = 3 and r = 5 we will have a permutation [1, 3, 2] for m = 3; 
  * if l = 1 and r = 5 we will have a permutation [4, 5, 1, 3, 2] for m = 5; 
  * if l = 1 and r = 6 we will have a permutation [4, 5, 1, 3, 2, 6] for m = 6; 
  * it is impossible to take some l and r, such that [p_l, p_{l+1}, …, p_r] is a permutation of numbers 1, 2, …, m for m = 2 and for m = 4. 



You are given a permutation p=[p_1, p_2, …, p_n]. For all m (1 ≤ m ≤ n) determine if it is a beautiful number or not.

Input

The first line contains the only integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. The next lines contain the description of test cases.

The first line of a test case contains a number n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the given permutation p. The next line contains n integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n, all p_i are different) — the given permutation p.

It is guaranteed, that the sum of n from all test cases in the input doesn't exceed 2 ⋅ 10^5.

Output

Print t lines — the answers to test cases in the order they are given in the input. 

The answer to a test case is the string of length n, there the i-th character is equal to 1 if i is a beautiful number and is equal to 0 if i is not a beautiful number.

Example

Input


3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2


Output


101011
11111
1001

Note

The first test case is described in the problem statement.

In the second test case all numbers from 1 to 5 are beautiful:

  * if l = 3 and r = 3 we will have a permutation [1] for m = 1; 
  * if l = 3 and r = 4 we will have a permutation [1, 2] for m = 2; 
  * if l = 2 and r = 4 we will have a permutation [3, 1, 2] for m = 3; 
  * if l = 2 and r = 5 we will have a permutation [3, 1, 2, 4] for m = 4; 
  * if l = 1 and r = 5 we will have a permutation [5, 3, 1, 2, 4] for m = 5. 
import sys
from math import gcd
from heapq import heappop
from heapq import heappush
from heapq import heapify
from bisect import insort
from bisect import bisect_right
from bisect import bisect_left
from sys import stdin,stdout
from collections import defaultdict, deque
from math import log2, ceil, floor

inp=lambda : int(input())
sip=lambda : input()
mulip =lambda : map(int,input().split())
lst=lambda : list(map(int,input().split()))
slst=lambda: list(sip())
arr2d= lambda x: [[int(j) for j in input().split()] for i in range(x)]
odds = lambda l: len(list(filter(lambda x: x%2!=0, l)))
evens = lambda l: len(list(filter(lambda x: x%2==0, l)))
mod = pow(10,9)+7
#-------------------------------------------------------
Judge = 0
if Judge:
    sys.stdin = open("input.in",'r')
    #sys.stdout = open("output.in",'w')

for _ in range(inp()):
    n = inp()
    p = lst()
    d = defaultdict(int)
    for i in range(n):
        d[p[i]]=i 
    d1 = {}
    l = []
    mn = n+1
    mx = -1
    for i in range(1,n+1):
        mn = min(mn,d[i])
        mx = max(mx,d[i])
        if i==1:
            print(1,end="")
            l.append(d[i]) 
        elif i==n:
            print(1,end="")
            l.append(d[i])
        else:
            l.append(d[i])
            if mx-mn+1==len(l):
                print(1,end="")
            else:
                print(0,end="")
    print()
    
    





