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;
template <class T>
vector<T> iarr(int n) {
  vector<T> a;
  while (n--) {
    T _v;
    cin >> _v, a.push_back(_v);
  }
  return a;
}
template <class T>
vector<vector<T>> imat(int n, int m) {
  vector<vector<T>> ma(n, vector<T>(m));
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> ma[i][j];
  return ma;
}
void solve();
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  int tt;
  cin >> tt;
  for (; tt-- > 0;) solve();
  return 0;
}
class BIT {
 public:
  BIT(int n = 0) { a = vector<int>(n + 1, 0); }
  int get(int i) {
    int s = 0;
    for (; i > 0; i -= i & -i) s += a[i];
    return s;
  }
  void set(int i, int v) {
    for (; i < a.size(); i += i & -i) a[i] += v;
  }
  vector<int> a;
};
void solve() {
  int n;
  cin >> n;
  auto a = iarr<int>(n);
  int j = -1;
  int v = 2;
  BIT b(n);
  for (int i = 0; i < n; i++) {
    if (a[i] == 1) {
      j = i;
      break;
    }
  }
  int l = j - 1, r = j + 1;
  b.set(1, 1);
  cout << 1;
  while (l >= 0 || r < n) {
    int ll = l >= 0 ? a[l] : 10000000;
    int rr = r < n ? a[r] : 10000000;
    if (ll < rr) {
      b.set(ll, 1);
      l--;
    } else {
      b.set(rr, 1);
      r++;
    }
    if (b.get(v) == v) {
      cout << 1;
    } else {
      cout << 0;
    }
    v++;
  }
  cout << "\n";
}
