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.*;
import java.io.*;

public class Solution{

	public static Integer INT(String s) {
		return Integer.parseInt(s);
	}
	public static Long LONG(String s) {
		return Long.parseLong(s);
	}

	static int mod=1_000_000_007, oo=Integer.MAX_VALUE, _oo=Integer.MIN_VALUE;
	static BufferedReader br=new BufferedReader(new InputStreamReader(System.in));	
	static Scanner in=new Scanner(System.in); 
	static StringBuilder out=new StringBuilder();

	//==================================================================================================================================================


	public static void main(String args[]) throws IOException {

		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 index;
			for(index=0; index<n; index++)
				if(a[index]==1)
					break;

			int lo=index, hi=index, max=1;
			int ans[]=new int[n+1];

			while(true) {

				if((hi-lo+1)==max)
					ans[hi-lo+1]=1;

				if(lo==0 && hi==n-1)
					break;

				if(lo==0)
					hi+=1;
				else if(hi==n-1)
					lo-=1;
				else {
					if(a[lo-1]<a[hi+1])
						lo-=1;
					else
						hi+=1;
				}

				max=Math.max(max, Math.max(a[lo], a[hi]));
			}

			for(int i=1; i<=n; i++)
				out.append(ans[i]);
			out.append("\n");
		}

		System.out.print(out);
	}
}