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;
string en = "\n";
string sp = " ";
string t = "hello";
string Y = "YES\n";
string N = "NO\n";
bool isprm(long long a) {
  for (long long i = 2; i * i <= a; i++) {
    if (a % i == 0) return false;
  }
  return true;
}
bool palin(string a) {
  for (int i = 0; i < a.size(); i++) {
    if (a.at(i) != a.at(a.size() - i - 1)) return 0;
  }
  return 1;
}
bool isVowel(char ch) {
  ch = toupper(ch);
  if (ch == 'A' || ch == 'U' || ch == 'I' || ch == 'O' || ch == 'E')
    return true;
  return false;
}
int toInt(string s) {
  int sm;
  stringstream ss(s);
  ss >> sm;
  return sm;
}
long long ceil2(long long a, long long b) {
  long long c = a / b;
  if (a % b) c++;
  return c;
}
long long bpow(long long a, long long b) {
  if (b == 0) return 1;
  long long r = bpow(a, b / 2);
  if (b % 2)
    return r * r * a;
  else
    return r * r;
}
long long binpow(long long a, long long b) {
  long long r = 1;
  while (b > 0) {
    if (b & 1) r = r * a;
    a = a * a;
    b >>= 1;
  }
  return r;
}
long long binpow(long long a, long long b, long long m) {
  a %= m;
  long long r = 1;
  while (b > 0) {
    if (b & 1) r = r * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return r;
}
template <typename T>
inline T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <typename T>
inline T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / gcd(a, b));
}
bool ar[200005];
int main() {
  long long i, j, k, n, m, l, s = 0, x, y, tc = 1, p, q, a, b, c = 0;
  bool f = 0, ff = 0;
  string st, s1, s2, s3;
  char ch;
  cin >> tc;
  while (tc--) {
    cin >> n;
    memset(ar, 0, sizeof(ar));
    map<long long, long long> mp;
    for (i = 0; i < n; i++) {
      cin >> x;
      mp[x] = i + 1;
    }
    ar[1] = ar[n] = 1;
    for (auto it = mp.begin(); it != mp.end(); it++) {
      x = it->first;
      if (x == 1)
        a = b = mp[x];
      else {
        if (mp[x] > a)
          b = max(b, mp[x]);
        else
          a = min(a, mp[x]);
      }
      if (b - a + 1 == x) ar[x] = 1;
    }
    for (i = 1; i <= n; i++) cout << ar[i];
    cout << en;
  }
}
