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 sys
input = sys.stdin.readline

class Bit:
    def __init__(self, n):
        self.size = n
        self.tree = [0] * (n + 1)
    
    def range_sum(self, l, r):
        return self.sum(r) - self.sum(l - 1)
        
        
    def sum(self, i):
        i += 1
        s = 0
        while i > 0:
            s += self.tree[i]
            i -= i & -i
        return s
        
    def get(self, i):
        return self.sum(i) - self.sum(i - 1)
 
    def add(self, i, x):
        i += 1
        while i <= self.size:
            self.tree[i] += x
            i += i & -i
 

def main():
    n = int(input())
    alst = list(map(int, input().split()))
    bit1 = Bit(n + 1)
    bit2 = Bit(n + 1)
    for a in alst:
        bit1.add(a, 1)
    ans = []
    l = 0
    r = n - 1
    pos = 1
    while r != l:
        if bit1.get(pos) >= 1 and bit1.sum(pos - 1) == 0 and bit2.sum(pos - 1) == pos - 1:
            ans.append("1")
        else:
            ans += ["0"] * (n - len(ans) - 1)
            break
        if alst[l] <= alst[r]:
            num = alst[l]
            l += 1
        else:
            num = alst[r]
            r -= 1
        bit1.add(num, -1)
        if bit2.get(num) == 0:
            bit2.add(num, 1)
        pos += 1
    lst = [i for i in range(1, n + 1)]
    alst.sort()
    for a, b in zip(alst, lst):
        if a != b:
            ans.append("0")
            print("".join(ans[::-1]))
            return
    ans.append("1")
    print("".join(ans[::-1]))
    
    
for _ in range(int(input())):
    main()