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.awt.Desktop;


import java.io.BufferedReader;


import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.net.URI;
import java.net.URISyntaxException;
import java.sql.Array;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.LinkedHashMap;
import java.util.LinkedHashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeSet;
import java.util.Vector;



public  class codechef3 {
static class comp implements Comparator<Integer>
{

	@Override
	public int compare(Integer o1, Integer o2) {
		if(Math.abs(o1)>Math.abs(o2))
			return -1;
		else return 1;
	
	}
	
	}
      //=======================================================
	  //sorting Pair
   static class comp1 implements Comparator<Pair<Integer,Integer>>
   {

	@Override
	public int compare(Pair<Integer, Integer> o1, Pair<Integer, Integer> o2) {
		if(o1.k>o2.k)
			return 1;
		else return -1;
	
	}

	
	
	}
        //======================================================= 
	   
   //Creating Pair class  
         //----------------------------------------------------------------------
	    
       static class Pair<Integer,Intetger>
       {
    	   int k=0;
    	   int v=0;
    	   public Pair(int a,int b)
    	   {
    		   k=a;
    		   v=b;
    	   }
    	   
    	   
    	   
       }
    
     //--------------------------------------------------------------------------
	
		 static class FastReader
		    {BufferedReader br; 
		        StringTokenizer st; 
		 public FastReader() 
		        {  br = new BufferedReader(new
		                     InputStreamReader(System.in)); 
		        } 
		  
		       String next() 
		        { 
		            while (st == null || !st.hasMoreElements()) 
		            { 
		                try
		                { 
		                    st = new StringTokenizer(br.readLine()); 
		                } 
		                catch (IOException  e) 
		                { 
		                    e.printStackTrace(); 
		                } 
		            } 
		            return st.nextToken(); 
		        } 
		  
		        int nextInt() 
		        { 
		            return Integer.parseInt(next()); 
		        } 
		  
		        long nextLong() 
		        { 
		            return Long.parseLong(next()); 
		        } 
		  
		        double nextDouble() 
		        { 
		            return Double.parseDouble(next()); 
		        } 
		  
		        String nextLine() 
		        { 
		            String str = ""; 
		            try
		            { 
		                str = br.readLine(); 
		            } 
		            catch (IOException e) 
		            { 
		                e.printStackTrace(); 
		            } 
		            return str; 
		        } 
		    } 
		
		 //gcd of two number
		 
		    public static int gcd(int a, int b)
		    {
		      if (b == 0)
		        return a;
		      return gcd(b, a % b); 
		    }
		    
		 //--------------------------------------------------------------------------------------------
		 
		 //lcm of two number
		 static int x;
		    public static int lcm(int a, int b)
		    {
		        return (a / gcd(a, b)) * b;
		    }
		 
		  //-------------------------------------------------------------------------------------------
		   
		 
		    
		    public static void main(String[] args) 
		    {
		    	FastReader s=new FastReader();
				int t=s.nextInt();
				while(t-->0)
				{
				int n=s.nextInt();
				int[] a=new int[n+1];
				
				for(int i=0;i<n;i++)
				{
					int num=s.nextInt();
					a[num]=i+1;
				}
				
				int min=Integer.MAX_VALUE;
				int max=Integer.MIN_VALUE;
				
				for(int i=1;i<=n;i++)
				{
					min=Math.min(min,a[i]);
					max=Math.max(max, a[i]);
					if(max-min==i-1)
						System.out.print("1");
					else System.out.print("0");
				}
				System.out.println();
					
					
				}
		    	
		    	
		    
		    	
		    	
		    	
}}