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 sys import stdin


class min_segmenttree:
    def __init__(self, arr, n):
        self.tree, self.n = [0] * (2 * n), n
        # build tree
        for i in range(2 * n - 1, 0, -1):
            if i >= n:
                self.tree[i] = arr[i - n]
            else:
                self.tree[i] = min(self.tree[i << 1], self.tree[(i << 1) + 1])

    # get interval[l,r)
    def query(self, l, r):
        res = float('inf')
        l += self.n
        r += self.n
        while l < r:
            if l & 1:
                res = min(self.tree[l], res)
                l += 1
            if r & 1:
                r -= 1
                res = min(self.tree[r], res)
            l >>= 1
            r >>= 1

        return res


rint = lambda: int(stdin.readline())
rints = lambda: [int(x) for x in stdin.readline().split()]
out = []
for _ in range(int(input())):
    n, a = rint(), rints()
    ans, tree = ['0'] * n, min_segmenttree(a, n)
    if tree.query(0, n) == 1:
        l, r, ans[-1] = 0, n - 1, '1'

        for i in range(1, n - 1):
            if a[l] != i and a[r] != i or a[l] == a[r] == i:
                break
            elif a[l] == i:
                l += 1
            else:
                r -= 1

            if tree.query(l, r + 1) != i + 1:
                break
            ans[n - i - 1] = '1'

    if len(set(a)) == n:
        ans[0] = '1'

    out.append(''.join(ans))

print('\n'.join(out))
