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;
long long binpow(long long a, long long b) {
  if (b == 0) return 1;
  long long res = binpow(a, b / 2);
  res = (res * res) % 1000000007LL;
  if (b % 2) {
    res = (res * a) % 1000000007LL;
  }
  return res;
}
long long inv(long long n) { return binpow(n, 1000000007LL - 2); }
bool compare(const pair<long long, long long>& i,
             const pair<long long, long long>& j) {
  if (i.first != j.first)
    return i.first < j.first;
  else
    return i.second > j.second;
}
vector<long long> factorize(long long n) {
  vector<long long> v;
  for (long long i = 2; i * i <= n; i++) {
    while (n % i == 0) {
      v.push_back(i);
      n /= i;
    }
  }
  if (n != 1) v.push_back(n);
  return v;
}
long long nCrModpDP(long long n, long long r) {
  long long C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = min(i, r); j > 0; j--)
      C[j] = (C[j] + C[j - 1]) % 1000000007LL;
  }
  return C[r];
}
long long binomialCoefficient(long long n, long long r) {
  if (r == 0) return 1;
  long long ni = n % 1000000007LL, ri = r % 1000000007LL;
  return (binomialCoefficient(n / 1000000007LL, r / 1000000007LL) *
          nCrModpDP(ni, ri)) %
         1000000007LL;
}
vector<bool> sieve(long long n) {
  vector<bool> PRIME(n + 1, true);
  PRIME[0] = false;
  PRIME[1] = false;
  for (long long p = 2; p * p <= n; p++) {
    if (PRIME[p] == true) {
      for (long long i = p * p; i <= n; i += p) {
        PRIME[i] = false;
      }
    }
  }
  return PRIME;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n, a;
    vector<pair<long long, long long> > v;
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> a;
      v.push_back({a, i});
    }
    sort((v).begin(), (v).end());
    long long maxm = v[0].second, minm = v[0].second;
    for (long long i = 0; i < n; i++) {
      maxm = max(maxm, v[i].second);
      minm = min(minm, v[i].second);
      if (maxm - minm == i)
        cout << "1";
      else
        cout << "0";
    }
    cout << "\n";
  }
}
