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.*;

import java.util.*;
 public class Solution {
	 
	
	public static void main(String[] args) throws Exception
    {
		FastReader fr=new FastReader();
	    int t=fr.nextInt();
	     while(t-->0) {
	    	int n=fr.nextInt();
	    	HashMap<Integer,Integer> map=new HashMap<>();
	    	for(int i=0;i<n;i++) {
	    		int v=fr.nextInt();
	    		map.put(v,i);
	    	}
	    	
	    	int min=n;int max=0;
	    	for(int i=1;i<=n;i++) {
	    		int id=map.get(i);
	    		
	    			min=Math.min(min, id);
	    			max=Math.max(max, id);
	    			if(max-min+1==i)
	    				System.out.print("1");
	    			else
	    				System.out.print("0");
	    		
	    	}
	    	
	    	System.out.println();
	    	
	        
	     }
    }
	
	
	
	 
public static String lcs(String a,String b) {
		int dp[][]=new int[a.length()+1][b.length()+1];
		for(int i=0;i<=a.length();i++)
		{
			for(int j=0;j<=b.length();j++)
			{
				if(i==0||j==0)
					dp[i][j]=0;
			}
		}
		
		for(int i=1;i<=a.length();i++)
		{
			for(int j=1;j<=b.length();j++)
			{
				if(a.charAt(i-1)==b.charAt(j-1))
					dp[i][j]=1+dp[i-1][j-1];
				else
					dp[i][j]=Math.max(dp[i-1][j],dp[i][j-1]);
			}
		}
		
		int i=a.length();
		int j=b.length();
		String lcs="";
		while(i>0&&j>0)
		{
			if(a.charAt(i-1)==b.charAt(j-1)) {
				lcs=a.charAt(i-1)+lcs;
				i--;
				j--;
			}
			else 
			{
				if(dp[i-1][j]>dp[i][j-1])
					i--;
				else
					j--;
			}
		}
		
		return lcs;
            
	}
	
	 
	 
	 
	 public static long facto(long n) {
		 if(n==1||n==0)
			 return 1;
		 return n*facto(n-1);
	 }
	 public static long gcd(long n1,long n2) {
		 if (n2 == 0) {
		        return n1;
		    }
		    return gcd(n2, n1 % n2);
	}
	 public static boolean isPali(String s) {
		 int i=0;
		 int j=s.length()-1;
		 while(i<=j) {
			 if(s.charAt(i)!=s.charAt(j))
				 return false;
			 i++;
			 j--;
		 }
		 return true;
	 }
	 public static String reverse(String s) {
		 String res="";
		 for(int i=0;i<s.length();i++) {
			 res+=s.charAt(i);
		 }
		 return res;
	 }
	
		
	 
	 public static int bsearch(long suf[],long val) {
		 int i=0;
		 int j=suf.length-1;
		 while(i<=j) {
			 int mid=(i+j)/2;
			 if(suf[mid]==val)
				 return mid;
			 else if(suf[mid]<val)
				 j=mid-1;
			 else
				 i=mid+1;
		 }
		 return -1;
	 }
	 
	 
	 
	
	 
	 
	public static int[] getFreq(String s) {
		int a[]=new int[26];
		for(int i=0;i<s.length();i++) {
			a[s.charAt(i)-'a']++;
		}
		return a;
		
	}
	 
	 public static boolean isPrime(int n) {
		 
		 for(int i=2;(i*i)<=n;i++) {
			 if(n%i==0)
				 return false;
		 }
		 return true;
	 }
            	  
             
             
    }
class Pair implements Comparable<Pair>{
	int start;
	int end;
	Pair(int start,int end){
		this.start=start;
		this.end=end;
	}
	public int compareTo(Pair o) {
		return this.start-o.start;
	}
	
}

	
	 
	
	

	 
 
  

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;
    }
}

