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. 
from collections import Counter


def solve():
    n = int(input())
    arr = [int(x) for x in input().split()]
    arr_count = Counter(arr)
    is_perm = '1'
    for i in range(1, n + 1):
        if arr_count[i] != 1:
            is_perm = '0'

    curr_l = 0
    curr_r = n - 1

    x___pos = [set() for _ in range(n + 1)]
    for i, x in enumerate(arr):
        x___pos[x].add(i)

    for x in range(1, n + 1):
        pos = x___pos[x]

        if len(pos) == 0:
            ans = ['1' for _ in range(x - 1)] + ['0' for _ in range(n)]
            ans = ans[:n - 1]
            ans.append(is_perm)
            ans.reverse()
            print(''.join(ans))
            return

        if len(pos) > 1:
            ans = ['1' for _ in range(x)] + ['0' for _ in range(n)]
            ans = ans[:n - 1]
            ans.append(is_perm)
            ans.reverse()
            print(''.join(ans))
            return

        pos = pos.pop()
        if pos != curr_l and pos != curr_r:
            ans = ['1' for _ in range(x)] + ['0' for _ in range(n)]
            ans = ans[:n - 1]
            ans.append(is_perm)
            ans.reverse()
            print(''.join(ans))
            return

        if pos == curr_l:
            curr_l += 1
        else:
            curr_r -= 1

    ans = [is_perm] + ['1' for _ in range(n - 1)]
    print(''.join(ans))


if __name__ == '__main__':
    for _ in range(int(input())):
        solve()
