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>
#pragma GCC optimize("O2")
using namespace std;
const double PI = atan(1.0) * 4;
const int di[4] = {-1, 0, 1, 0};
const int dj[4] = {0, -1, 0, 1};
const long long INF = (long long)2e18 + 50;
const int maximum = numeric_limits<int>::min();
const int minimum = numeric_limits<int>::max();
const long long mod = 1e9 + 7;
long long power(long long a, long long e) {
  long long res = 1LL;
  while (e > 0) {
    if (e % 2 == 1) res = res * a;
    a = a * a;
    e /= 2;
  }
  return res;
}
bool isBalanced(const string &exp) {
  bool flag = true;
  int count = 0;
  for (int i = 0; i < exp.length(); i++) {
    if (exp[i] == '(') {
      count++;
    } else {
      count--;
    }
    if (count < 0) {
      flag = false;
      break;
    }
  }
  if (count != 0) {
    flag = false;
  }
  return flag;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> p(n);
    vector<int> pos(n);
    for (int i = 0; i < n; i++) {
      cin >> p[i];
      --p[i];
      pos[p[i]] = i;
    }
    int pos_max = -1, pos_min = n + 1;
    vector<int> res(n);
    for (int i = 0; i < n; i++) {
      pos_max = max(pos_max, pos[i]);
      pos_min = min(pos_min, pos[i]);
      if (pos_max - pos_min == i)
        res[i] = 1;
      else
        res[i] = 0;
    }
    for (auto i : res) {
      cout << i;
    }
    cout << "\n";
  }
  return 0;
}
