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;
struct pair_hash {
  template <class T1, class T2>
  std::size_t operator()(const std::pair<T1, T2>& pair) const {
    return std::hash<T1>()(pair.first) ^ std::hash<T2>()(pair.second);
  }
};
const long double EPS = 1e-7;
const long long MOD = 1000000007;
const long long INF = 1e18;
const double pi = 2 * acos(0.0);
long long n, m, k, p, q, t, sum = 0, cnt = 0;
long long mx = INT64_MIN;
long long mn = INT64_MAX;
bool flag = 0;
vector<long long> v;
string s = "", s1, s2;
void bad_solve() {}
void solve() {
  cin >> t;
  while (t--) {
    cin >> n;
    vector<long long> pos(n + 1, 0);
    v.clear();
    for (long long(i) = 0; (i) < (n); ++(i)) {
      cin >> p;
      pos[p] = i;
      v.push_back(p);
    }
    vector<bool> ans(n, 0);
    long long l = pos[1];
    long long r = pos[1];
    ans[0] = 1;
    sum = 1;
    for (long long(i) = (2); (i) < (n + 1); ++(i)) {
      if (pos[i] < l) {
        while (l != pos[i]) --l;
      } else if (pos[i] > r) {
        while (r != pos[i]) ++r;
      }
      if (r - l + 1 == i) ans[i - 1] = 1;
    }
    {
      long long length_of_vector = ans.size();
      for (long long(elementvec) = 0; (elementvec) < (length_of_vector);
           ++(elementvec)) {
        cout << ans[elementvec] << "";
      }
    };
    cout << endl;
  }
}
int main() {
  cout.precision(20);
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
