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 Main {
	public static void main (String[] args) {
		Scanner in = new Scanner(System.in);
		
		int t = in.nextInt();
		while(t-->0) {
		    int n = in.nextInt();
		    int a[] = new int[n];
		    for(int i=0;i<n;i++)a[i] = in.nextInt();
		    int ind[] = new int[n+1];
		    
		    for(int i=0; i<n; i++) {
		        ind[a[i]] = i+1;
		    }
		    int ans[] = new int[n+1];
		    ans[1] = 1; ans[n] = 1;
		    if(n==1) System.out.println(1);
		    else {
		        if(Math.abs(ind[2]-ind[1]) == 1) ans[2]=1;
		        else ans[2]=0;
    		    int min = Math.min(ind[1],ind[2]);
    		    int max = Math.max(ind[1],ind[2]);
    		    for(int i=3; i<n+1; i++) {
    		        if(max-min >= i || max-ind[i]>=i || ind[i]-min>=i) ans[i] = 0;
    		        else ans[i] = 1;
    		        min = Math.min(min, ind[i]);
    		        max = Math.max(max, ind[i]);
    		    }
    		    for(int i=1; i<n+1; i++) System.out.print(ans[i]+"");
    		    System.out.println();
		    }
		}
		
	}
}