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. 
import java.util.*;
public class Q2 {
    static int m = 1000000007;
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int t = s.nextInt();
        long fact[] = new long[200001];
        fact[0] = 1;
        fact[1] = 1;
        for(int i = 2 ; i <= 200000 ; i++)
            fact[i] = (i*fact[i-1])%m;
        while(t-->0){
            int n = s.nextInt();

            long num[] = new long[n];
            int l = 0 , r = 0;

            for(int i = 0 ; i < n ; i++)
            {
                num[i] = s.nextLong();
                if(num[i] == 1)
                {
                    l = i;
                    r = i;
                }
            }

            int[] ans = new int[n];

            ans[0] = 1;
            long sum = 1;
            if(n == 1){
                System.out.println(1);
                continue;
            }

            while (l != 0 || r != n - 1){
                if (l > 0 && r < n - 1) {
                    if (num[l - 1] < num[r + 1]) {
                        sum = (sum%m*num[l - 1]%m)%m;
                        l--;
                    } else {
                        sum = (sum%m*num[r+1]%m)%m;
                        r++;
                    }

                    //long lol = (r - l + 1) * (r - l + 2) / 2;
                    if (sum == fact[r-l+1])
                        ans[r - l] = 1;
                } else if (l == 0) {
                    sum = (sum%m*num[r+1]%m)%m;
                    r++;

                    //long lol = (r - l + 1) * (r - l + 2) / 2;
                    if (sum == fact[r-l+1])
                        ans[r - l] = 1;
                } else if (r == n - 1) {
                    sum = (sum%m*num[l- 1]%m)%m;
                    l--;

                    //long lol = (r - l + 1) * (r - l + 2) / 2;
                    if (sum == fact[r-l+1])
                        ans[r - l] = 1;
                }

            }

            for(int i = 0 ; i < n ; i++)
                System.out.print(ans[i]);

            System.out.println();
        }
    }
}