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.Scanner;

/*
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 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */

/**
 *
 * @author zye
 */
public class B {
    public static void main(String args[]) {
        Scanner scanner = new Scanner(System.in);
        
        int t = scanner.nextInt();
        for (int i = 0; i < t; i++) {
            int n = scanner.nextInt();
            int[] permutation = new int[n];
            for (int j = 0; j < n; j++) {
                permutation[j] = scanner.nextInt();
            }
            System.out.println(toString(isBeautiful(permutation)));
        }
    }
    
    public static String toString(boolean[] booleans) {
        StringBuilder str = new StringBuilder();
        for (boolean b : booleans) {
            str.append(b ? '1' : '0');
        }
        return str.toString();
    }
    
    public static boolean[] isBeautiful(int[] permutation) {
        boolean[] marks = new boolean[permutation.length];
        int pos[] = new int[permutation.length + 1];
        for (int i = 0; i < permutation.length; i++) {
            pos[permutation[i]] = i;
        }
        int min = pos[1], max = pos[1];
        marks[0] = true;
        for (int i = 2; i <= permutation.length; i++) {
            min = Math.min(min, pos[i]);
            max = Math.max(max, pos[i]);
            marks[i - 1] = max - min + 1 == i;
                
        }
        return marks;
    }
}
