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.*;
import java.lang.reflect.Array;
    	    	
 public class notYet {
 
    public static String sortString(String inputString) 
    { 
        // convert input string to char array 
        char tempArray[] = inputString.toCharArray(); 
          
        // sort tempArray 
        Arrays.sort(tempArray); 
          
        // return new sorted string 
        return new String(tempArray); 
    } 
	
    	  
	public static void main(String[] args) throws IOException
	{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		
		int t = Integer.parseInt(br.readLine());
		
		while(t-- > 0)
		{
			int n = Integer.parseInt(br.readLine());
			
			pair arr[] = new pair[n];
			
			StringTokenizer st1 = new StringTokenizer(br.readLine());
			
			

			StringBuilder sb = new StringBuilder("1");
			for(int i = 0 ; i < n ; i++)
			{
				int tmp = Integer.parseInt(st1.nextToken());
				
				arr[i] = new pair(tmp, i);
				
			
			}
			
			Arrays.sort(arr);
			
			int minInd = arr[0].y;
			int maxInd = arr[0].y;
			
			for(int i = 1 ; i < n ; i++)
			{
				if((arr[i].y > minInd && arr[i].y > maxInd)|| (arr[i].y < minInd && arr[i].y< maxInd))
				{
					int s = Math.abs(arr[i].y - maxInd) + Math.abs(arr[i].y - minInd);
					
					if(s == i+1)
					{
						sb.append(1);
					}
					else
					{
						sb.append(0);
					}
				}
				else
				{
					int s = Math.abs(arr[i].y - maxInd) + Math.abs(arr[i].y - minInd);
					
					if(s == i)
					{
						sb.append(1);
					}
					else
					{
						sb.append(0);
					}
				}
				
				minInd = Math.min(minInd, arr[i].y);
				maxInd = Math.max(maxInd, arr[i].y);
				
				
			}
			
			out.println(sb);
			
		
					
			
			
			
		}
		
		out.flush();
		out.close();
	}
	
	static class pair implements Comparable{
		
		int x , y;
		
		public pair(int x , int y)
		{
			this.x = x;
			this.y = y;
		}

		@Override
		public int compareTo(Object o) {
			pair p = (pair)o;
			
			if(this.x > p.x)
				return 1;
			if(this.x < p.x)
				return -1;
			return 0;
			
		}
		
	}
    		
    		
    		
    	
  }