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. 
#include <bits/stdc++.h>
using namespace std;
void Fast();
long long mindig(long long x);
long long maxdig(long long x);
long long bpow(long long a, long long b, long long abs);
long long bis(long long ar[], long long l, long long r, long long x);
long long bpow(long long a, long long b, long long modi) {
  a %= modi;
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = res * a % modi;
    }
    a = a * a % modi;
    b >>= 1;
  }
  return res;
}
long long bis(long long ar[], long long l, long long r, long long x) {
  while (l <= r) {
    long long m = l + (r - l) / 2;
    if (ar[m] == x) {
      return m;
    }
    if (ar[m] < x) {
      l = m + 1;
    } else {
      r = m - 1;
    }
  }
  return -1;
}
long long mindig(long long x) {
  long long ans = 9;
  long long t = x;
  while (x) {
    ans = min(ans, x % 10);
    x /= 10;
  }
  x = t;
  return ans;
}
long long maxdig(long long x) {
  long long ans = 0;
  long long t = x;
  while (x) {
    ans = max(ans, x % 10);
    x /= 10;
  }
  x = t;
  return ans;
}
void Fast() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  Fast();
  long long q;
  cin >> q;
  while (q--) {
    long long n;
    cin >> n;
    long long ar[n];
    for (long long i = 0; i < n; i++) cin >> ar[i];
    ;
    long long l, r;
    map<long long, long long> m;
    for (long long i = 0; i < n; i++) {
      m[ar[i]] = i + 1;
    }
    cout << 1;
    l = m[1], r = m[1];
    for (long long i = 2; i < n + 1; i++) {
      l = min(l, m[i]);
      r = max(r, m[i]);
      if (r - l > i - 1)
        cout << 0;
      else
        cout << 1;
    }
    cout << "\n";
  }
}
