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 deque


def readGenerator():
    while True:
        tokens = input().split(' ')
        for t in tokens:
            yield t


reader = readGenerator()


def readWord():
    return next(reader)


def readInt():
    return int(next(reader))


def readFloat():
    return float(next(reader))


def readLine():
    return input()


def createCompression(a, n, k):
    m = deque()
    res = set()
    for i in range(n):
        if len(m) > 0 and m[0][1] < i - k:
            m.popleft()
        while len(m) > 0 and m[-1][0] >= a[i]:
            m.pop()

        m.append((a[i], i))
        if i - k > -1:
            res.add(m[0][0])

    return len(res) == n - k and max(res) == n - k


def solve(n, a):
    if n == 1:
        return '1' if a[0] == 1 else '0'

    if createCompression(a, n, 1) and createCompression(a, n, 0):
        return '1' * n
    if not createCompression(a, n, n - 1):
        return '0' * n

    s = '1' if createCompression(a, n, 0) else '0'
    l = 0
    r = n - 1
    while (r - l) > 1:
        mid = (r + l) // 2
        if createCompression(a, n, mid):
            r = mid
        else:
            l = mid

    if createCompression(a, n, l):
        return s + ('0' * (l - 1)) + ('1' * (n - l))
    return s + ('0' * (r - 1)) + ('1' * (n - r))


t = readInt()
for _ in range(t):
    n = readInt()
    a = [readInt() for _ in range(n)]
    print(solve(n, a))
