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. 
import io
import os

from collections import Counter, defaultdict, deque

# https://raw.githubusercontent.com/cheran-senthil/PyRival/master/pyrival/data_structures/RangeQuery.py
class RangeQuery:
    def __init__(self, data, func=min):
        self.func = func
        self._data = _data = [list(data)]
        i, n = 1, len(_data[0])
        while 2 * i <= n:
            prev = _data[-1]
            _data.append([func(prev[j], prev[j + i]) for j in range(n - 2 * i + 1)])
            i <<= 1

    def query(self, start, stop):
        """func of data[start, stop)"""
        depth = (stop - start).bit_length() - 1
        return self.func(
            self._data[depth][start], self._data[depth][stop - (1 << depth)]
        )

    def __getitem__(self, idx):
        return self._data[0][idx]


def isPerm(C):
    return len(set(C)) == len(C) and min(C) == 1 and max(C) == len(C)


def solveBrute(N, A):
    rmq = RangeQuery(A)
    ans = []
    for k in range(1, N + 1):
        C = []
        for i in range(N - k + 1):
            C.append(rmq.query(i, i + k))
        print(C)
        if isPerm(C):
            ans.append("1")
        else:
            ans.append("0")

    return "".join(ans)


def solve(N, A):
    ans = ["0"] * N

    # k = 1 case
    if isPerm(A):
        ans[1 - 1] = "1"
    else:
        ans[1 - 1] = "0"
    # k = N case
    if min(A) == 1:
        ans[N - 1] = "1"
    else:
        ans[N - 1] = "0"

    rmq = RangeQuery(A)

    indices = defaultdict(list)
    for i, x in enumerate(A):
        indices[x].append(i)


    i = 0
    j = N - 1
    for x in range(1, N + 1):
        mnIndices = indices[x]
        if len(mnIndices) == 1 and mnIndices[0] in [i, j]:
            if mnIndices[0] == i:
                i += 1
            else:
                assert mnIndices[0] == j
                j -= 1
            ans[-x] = "1"
        elif rmq.query(i, j + 1) == x:
            ans[-x] = "1"
            break
        else:
            break

    return "".join(ans)


DEBUG = False
if DEBUG:
    import random

    random.seed(0)
    for _ in range(10000):
        N = random.randint(1, 10)
        A = [random.randint(1, 10) for i in range(N)]
        print("tc", _, N, A)
        ans1 = solve(N, A)
        ans2 = solveBrute(N, A)
        print(ans1, ans2)
        assert ans1 == ans2
    exit()

if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    TC = int(input())
    for tc in range(1, TC + 1):
        (N,) = [int(x) for x in input().split()]
        A = [int(x) for x in input().split()]
        ans = solve(N, A)
        print(ans)
