You are given a permutation p=[p_1, p_2, …, p_n] of integers from 1 to n. Let's call the number m (1 ≤ m ≤ n) beautiful, if there exists two indices l, r (1 ≤ l ≤ r ≤ n), such that the numbers [p_l, p_{l+1}, …, p_r] is a permutation of numbers 1, 2, …, m.

For example, let p = [4, 5, 1, 3, 2, 6]. In this case, the numbers 1, 3, 5, 6 are beautiful and 2, 4 are not. It is because:

  * if l = 3 and r = 3 we will have a permutation [1] for m = 1; 
  * if l = 3 and r = 5 we will have a permutation [1, 3, 2] for m = 3; 
  * if l = 1 and r = 5 we will have a permutation [4, 5, 1, 3, 2] for m = 5; 
  * if l = 1 and r = 6 we will have a permutation [4, 5, 1, 3, 2, 6] for m = 6; 
  * it is impossible to take some l and r, such that [p_l, p_{l+1}, …, p_r] is a permutation of numbers 1, 2, …, m for m = 2 and for m = 4. 



You are given a permutation p=[p_1, p_2, …, p_n]. For all m (1 ≤ m ≤ n) determine if it is a beautiful number or not.

Input

The first line contains the only integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. The next lines contain the description of test cases.

The first line of a test case contains a number n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the given permutation p. The next line contains n integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n, all p_i are different) — the given permutation p.

It is guaranteed, that the sum of n from all test cases in the input doesn't exceed 2 ⋅ 10^5.

Output

Print t lines — the answers to test cases in the order they are given in the input. 

The answer to a test case is the string of length n, there the i-th character is equal to 1 if i is a beautiful number and is equal to 0 if i is not a beautiful number.

Example

Input


3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2


Output


101011
11111
1001

Note

The first test case is described in the problem statement.

In the second test case all numbers from 1 to 5 are beautiful:

  * if l = 3 and r = 3 we will have a permutation [1] for m = 1; 
  * if l = 3 and r = 4 we will have a permutation [1, 2] for m = 2; 
  * if l = 2 and r = 4 we will have a permutation [3, 1, 2] for m = 3; 
  * if l = 2 and r = 5 we will have a permutation [3, 1, 2, 4] for m = 4; 
  * if l = 1 and r = 5 we will have a permutation [5, 3, 1, 2, 4] for m = 5. 
def sleft(ls: list, d: int):
    t = 0
    for i in ls[::-1]:
        if i < d:
            t += 1
        else:
            return t
    return t


def sright(ls: list, d: int):
    t = 0
    for i in ls:
        if i < d:
            t += 1
        else:
            return t
    return t


def search(s: list):
    if len(s) == 1:
        if s[0] == 1:
            return '1'
        else:
            return '0'
    l = len(s)
    ans = ['0' for i in range(l)]
    for i, num in enumerate(s):
        j = 0
        if i != 0:
            j += sleft(s[:i], num)
        if i != l-1:
            j += sright(s[i+1:], num)
        if j == num-1:
            ans[num-1] = '1'
        else:
            ans[num-1] = '0'
    return ''.join(ans)


def alt(s: set, lr: int):
    s = [-1] + sorted(s) + [lr]
    jd = 0
    for i in range(1, len(s)):
        if s[i] != s[i-1]+1:
            jd += 1
        if jd == 2:
            return False
    return True


def search2(s: list, l: int):
    if len(s) == 1:
        return '1'
    d = {s.index(l)}
    out = '1'
    for i in range(l-1, 0, -1):

        if alt(d, l):
            out += '1'
        else:
            out += '0'
        p = s.index(i)
        d.add(p)
        #print(d)
    return out[::-1]


def search3(s: list, l: int):
    if l == 1:
        return '1'
    if l == 2:
        return '11'
    out = '1'
    large = s.index(1)
    small = s.index(1)
    dic = sorted(enumerate(s), key=lambda x: x[1])[1:]
    for i in dic:
        mid = i[0]
        if mid > large:
            large = mid
        if mid < small:
            small = mid
        if large-small == i[1]-1:
            out += '1'
        else:
            out += '0'
    return out


n = int(input())
out = []
for i in range(n):
    a = int(input())
    l = list(map(int, input().split(' ')))
    out.append(search3(l, a))
for i in out:
    print(i)
'''
import random
a = list(range(1, 20))
random.shuffle(a)
print(a)
print(search3(a, len(a)))
'''
