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. 
def read_ints():
    return map(int, input().split())


t_n, = read_ints()
for i_t in range(t_n):
    n, = read_ints()
    a_seq = tuple(read_ints())
    result = [None] * n
    result[0] = sorted(a_seq) == list(range(1, n+1))
    a_need = 1
    i, j = 0, n - 1
    has_analogue = False
    while i <= j and (a_seq[i] == a_need or a_seq[j] == a_need):
        if a_seq[i] == a_need and a_seq[j] == a_need and i != j:
            i += 1
            a_need += 1
            has_analogue = True
            break
        else:
            if a_seq[i] == a_need:
                i += 1
            else:
                j -= 1
            a_need += 1
    wrong_segment_length = j - i + 1
    if not has_analogue:
        min_k = wrong_segment_length
        wrong_segment_min = min(a_seq[i:j+1], default=n+1)
    else:
        min_k = wrong_segment_length + 1
        wrong_segment_min = min(a_seq[i:j+1])
    if wrong_segment_min == a_need:
        min_k = max(min_k, wrong_segment_length)
    elif wrong_segment_min > a_need:
        min_k = max(min_k, wrong_segment_length + 1)
    else:
        while wrong_segment_min != a_need:
            a_need -= 1
            wrong_segment_length += 1
        min_k = max(min_k, wrong_segment_length)
    
    for k in range(min_k, n+1):
        result[k-1] = True
    print(''.join(['1' if result[i] else '0' for i in range(n)]))
