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 math
def check(k):
    a=1
    b=-1
    c=-k*2
    d=(b**2) -4*a*c
    x=(-1+math.sqrt(d))/(2*a)
    x=int(x)
    if (x*(x+1))/2==k:
        return x
    else:
        return 0
def solve(l,n):
    i=0
    while l[i]!=1:
        i+=1
    j=i+1
    i-=1
    s=1
    while (i>=0 or j<n):
        if (i>=0 and j<n):
            if l[i]<l[j]:
                s+=l[i]
                if check(s)==j-i:
                    ans[j-i]=1
                i-=1
            else:
                s+=l[j]
                if check(s)==j-i:
                    ans[j-i]=1
                j+=1
        else:
            if i>=0:
                s+=l[i]
                if check(s)==j-i:
                    ans[j-i]=1
                i-=1
            else:
                s+=l[j]
                if check(s)==j-i:
                    ans[j-i]=1
                j+=1
for _ in range(input()):
    n=input()
    ans=[0]*(n+1)
    ans[1]=1
    solve(map(int,raw_input().split()),n)
    s=''
    for i in range(1,len(ans)):
        s+=str(ans[i])
    print s