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 double EPS = 1e-8;
const long long mod = (1e9) + 7;
int dx[] = {0, 0, -1, 1, 1, 1, -1, -1};
int dy[] = {1, -1, 0, 0, 1, -1, -1, 1};
const int N = 1000001;
void stop() {}
void file() {}
struct point {
  long long x, y;
};
unsigned long long gcd(unsigned long long a, unsigned long long b) {
  return !b ? a : gcd(b, a % b);
}
long long fixMod(long long a, long long b) { return ((a % b) + b) % b; }
double triangle_area(int x1, int y1, int x2, int y2, int x3, int y3) {
  return abs((x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1)) / 2.;
};
long long dis(pair<long long, long long> a, pair<long long, long long> b) {
  long long x = a.first - b.first, y = a.second - b.second;
  return abs(x) + abs(y);
}
unsigned long long LCM(unsigned long long a, unsigned long long b) {
  return a / gcd(a, b) * b;
}
long long power(long long x, long long y) {
  if (y == 0) return 1;
  if (y == 1) return x;
  long long r = power(x, y / 2) % mod;
  return r % mod * r % mod * power(x, y % 2) % mod;
}
bool isPrime(long long x) {
  if (x <= 1) return 0;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> v(n);
    map<int, int> mp;
    for (int i = 0; i < n; i++) {
      cin >> v[i];
      mp[v[i]] = i + 1;
    }
    int mn = n + 1, mx = 0;
    for (int i = 0; i < n; i++) {
      mn = min(mp[i + 1], mn);
      mx = max(mp[i + 1], mx);
      if (mx - mn == i)
        cout << 1;
      else
        cout << 0;
    }
    cout << '\n';
  }
  return 0;
}
