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;
const long double eps = 1e-16;
const int MOD = (int)1e9 + 9;
const int INF = (int)1e9;
const int NHASH = 2;
template <typename T>
vector<T> readVector0(int n) {
  vector<T> res(n);
  for (int i = 0; i < n; i++) cin >> res[i];
  return res;
}
template <typename T>
vector<T> readVector1(int n) {
  vector<T> res(n + 1);
  for (int i = 1; i <= n; i++) cin >> res[i];
  return res;
}
inline long long mod(long long first, long long m = MOD) {
  if (first < 0) first += m;
  return (first % m);
}
template <typename T>
bool sortBy2(const pair<T, T> &a, const pair<T, T> &b) {
  return (a.second < b.second);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  vector<string> ans(t, "1");
  for (int k = 0; k < t; ++k) {
    int n;
    cin >> n;
    vector<int> a(n + 2, 0);
    vector<int> inv(n + 2, 0);
    a[0] = n + 1;
    a[n + 1] = n + 1;
    for (int i = 1; i <= n; ++i) {
      cin >> a[i];
      inv[a[i]] = i;
    }
    int l = inv[1];
    int r = inv[1];
    int mx = 1;
    for (int i = 2; i <= n; i++) {
      if (a[l - 1] < a[r + 1]) {
        l--;
        if (a[l] > mx) {
          mx = a[l];
        }
      } else {
        r++;
        if (a[r] > mx) {
          mx = a[r];
        }
      }
      if (r - l + 1 >= mx)
        ans[k] += '1';
      else
        ans[k] += '0';
    }
  }
  for (int k = 0; k < t; ++k) {
    cout << ans[k] << endl;
  }
  return 0;
}
