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.IOException;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.util.*;
import java.math.*;
import java.io.*;
import java.text.*;
import java.math.BigInteger;
 
public class Main
{  
 
    //Life's a bitch
    public static boolean[] sieve(long n)
    {
        boolean[] prime = new boolean[(int)n+1];
        Arrays.fill(prime,true);
        prime[0] = false;
        prime[1] = false;
        long m = (long)Math.sqrt(n);
        for(int i=2;i<=m;i++)
        {
            if(prime[i])
            {
                for(int k=i*i;k<=n;k+=i)
                {
                    prime[k] = false;
                }
            }
        }
        return prime;
    } 
 
    
    static long GCD(long a,long b)
    {
        if(b==0)
        {
            return a;
        }
        return GCD(b,a%b);
    }
 
    static long CountCoPrimes(long n)
    {
        long res = n;
        for(int i=2;i*i<=n;i++)
        {
            if(n%i==0)
            {
                while(n%i==0)
                {
                    n/=i;
                }
                res-=res/i;
            }
        }
        if(n>1)
        {
            res-=res/n;
        }
        return res;
    }
 
 
    
 
    
 
    static boolean prime(int n)
    {
        for(int i=2;i*i<=n;i++)
        {
            if(i%2==0 ||i%3==0)
            {
                return false;
            }
        }
        return true;
    }

    
 
   
 
 
   
   
  
    public static void main(String[] args) throws IOException
    {
     
        new Main().run();
       
    }
 
 	static void reverse(long[] a)
 	{	
 		int end = a.length-1;
 		int start = 0;
 		while(start<end)
 		{
 			long temp = a[start];
 			a[start] = a[end];
 			a[end] = temp;
 			start++;
 			end--;
 		}
 	}
   
 
    Scanner in = new Scanner(System.in);
    // static int[] arr,tree[],lev;
    void run() throws IOException
    {
    	int q = ni();
    	while(q-->0)
    	{
    		int n = ni();
    		int[] a = new int[n];
    		for(int i=0;i<n;i++)
    		{
    			int ai = ni();
    			a[ai-1] = i;
    		}

    		int min = n;
    		int max = -1;
    		for(int i=0;i<n;i++)
    		{
    			min = Math.min(min,a[i]);
    			max = Math.max(max,a[i]);
    			if(max-min==i)
    			{
    				System.out.print(1);
    			}
    			else{
    				System.out.print(0);
    			}
    		}

    		System.out.println();




    		
    	}
    	
		


   	}


   	

 	

   	
	


	
	


   

 

    
 
 
 
 
 
 
    
 
    //xor range query
    static long xor(long n)
    {
 
        if(n%4==0)
        {
            return n;
        }
        if(n%4==1)
        {
            return  1;
        }
        if(n%4==2)
        {
            return n+1;
        }
        return 0;
    }
 
    static long xor(long a,long b)
    {
        return xor(b)^xor(a-1);
    }
 
 
    
 
  
   
    
 
    
       
    
 
    
   
  
 
 
 
   
 
     void printL(long a)
    {
        System.out.println(a);
    }
    void printS(String s)
    {
        System.out.println(s);
    }
 
     void printD(Double d)
    {
        System.out.println(d);
    }
  
   
   
    static void swap(char c,char p)
    {
        char t = c;
        c = p;
        p = t;
    }
 
    static long max(long n,long m)
    {
        return Math.max(n,m);
    }
    static long min(long n,long m)
    {
        return Math.min(n,m);
    }
 
    double nd() throws IOException
    {
        return Double.parseDouble(in.next());
    }
    int ni() throws IOException
    {
        return Integer.parseInt(in.next());
    }
 
    long nl() throws IOException
    {
        return Long.parseLong(in.next());
    }
 
    String si() throws IOException
    {
        return in.next();
    }
 
 
  
   
    static int abs(int n)
    {
        return Math.abs(n);
    }
 
    static class Scanner 
    {
        StringTokenizer st;
        BufferedReader br;
 
        public Scanner(InputStream s){  br = new BufferedReader(new InputStreamReader(s));}
 
        public String next() throws IOException 
        {
            while (st == null || !st.hasMoreTokens()) 
            {
                st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
        }
 
        public int nextInt() throws IOException {return Integer.parseInt(next());}
        
        public long nextLong() throws IOException {return Long.parseLong(next());}
 
        public String nextLine() throws IOException {return br.readLine();}
 
        public boolean ready() throws IOException {return br.ready();}
 
 
    }
 
    
}
 
 
 
class Pair implements Comparable<Pair>
{
   int x,y;
    public Pair(int x,int y)
    {
        this.x = x;
        this.y = y;
    }
    public int compareTo(Pair o)
    {
        return this.y-o.y;
    }
}