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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.*;
import java.util.*;

public class B {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
// 		Scanner sc = new Scanner(System.in);
// 		String xy = sc.nextLine();
// 		 String[] numbers = xy.split(" ");
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String xy="";
try{xy = br.readLine();
}catch(IOException e){
	
}
int p[]=new int[200007];
int l[]=new int[200007];
		
int t=Integer.parseInt(xy);
while(t>0) {
	t--;
	try{xy = br.readLine();
	}catch(IOException e){}
	int n=Integer.parseInt(xy);
		try{xy = br.readLine();
	}catch(IOException e){}
	 String[] numbers = xy.split(" ");
	 for(int i=0;i<n;i++) {
		 p[i]=Integer.parseInt(numbers[i]);	
		 l[p[i]]=i;
}	
	int  b=l[1];
	int e=l[1];
System.out.print("1");	
int diff=0;

		

for(int i=2;i<=n;i++) {
//		if(b<=p[i]&&p[i]<=e) {
//				}
//		if(l[i]<b&&l[i]<e) {
//			b=l[i];
//		}
//		if(l[i]>b&&l[i]>e)			
//		e=l[i];
		b=Math.min(l[i], b);
		e=Math.max(l[i], e);
		//	System.out.println(l[i]+"  "+b+"  "+e);

		diff=e-b;
	 if(diff<i)
System.out.print("1");		 else
System.out.print("0");
	
	}
	
System.out.println();
	

	
}

	}

}



