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. 
//package practice_codes;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.stream.Collectors;

//import codes.Reference.i;

public class BeautifulNumbers {
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		
		long t;
		
		t = i.l();
		//String s = i.s();
		while(t>0)
		{
			long n = i.l();
			long a[]=i.la();
			long b[] = new long[a.length];
			for(int i=0;i<a.length;i++)
			{
				b[(int) (a[i]-1)] = i;
			}
			long l=a.length;
			long r = 0;
			for(int i=0;i<a.length;i++)
			{
				l = Math.min(l, b[i]);
				r = Math.max(r, b[i]);
				if(r-l==i)
					System.out.print("1");
				else
					System.out.print("0");
			}
			System.out.println();
			t--;	
		}
	}
	static private class i
	{
		//input taking methods
		static BufferedReader b = new BufferedReader(new InputStreamReader(System.in));
		
		public static String[] sa()
	    {
	    	try {
				return b.readLine().split(" ");
			} catch (IOException e) {
				
				e.printStackTrace();
			}
	    	return null;
	    }
		
		public static long l()
		{
			try {
				return Long.parseLong(b.readLine());
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
	    	return 0;
		}
		
		public static String s()
		{
			try {
				return b.readLine();
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
	    	return null;
		}
		
		public static long[] la()
		{
			return Arrays.stream(sa()).parallel().mapToLong(Long::parseLong).toArray();
		}
		
		public static ArrayList<Long> lal()
		{
			return (ArrayList<Long>)Arrays.stream(la()).parallel().boxed().collect(Collectors.toList());
		}
		public static StringBuilder so(String a)
		{
			return new StringBuilder(a);
		}

	}
	/*
	 HashMap<String,Integer> map = new HashMap<String,Integer>();
	 if(!map.containsKey(key))
	 {
	 	map.put(key,value);
	 }
	 else
	 {
	 	map.replace(key,map.get(key));
	 } 
	 */
	static private class algo{
		
		public static boolean isPrime(int a)
		{
			for(int i=2;i<a;i++)
			{
				if(a%i==0)
				{
					return false;
				}
			}
			return true;
		}
		static String reverse(String s)
		{
			return i.so(s).reverse().toString();
		}
		static String replace(String s,int st,int e,String repl)
		{
			return i.so(s).replace(st, e, repl).toString();
		}
		static String insert(String s,int pos,String ins)
		{
			return i.so(s).insert(pos, ins).toString();
		}
		static long sod(long n) 
	    { 
	        long sum; 
	        for (sum = 0; n > 0; sum += n % 10, 
	                                  n /= 10); 
	  
	        return sum; 
	    } 

		static int gcd(int a, int b) 
	    { 
			if (a == 0) 
				return b;  
			return gcd(b % a, a);  
	    } 
		
	    static int lcm(int a, int b) 
	    { 
	        return (a*b)/gcd(a, b); 
	    }
	    static long sumOfSquares(long[] a)
	    {
	    	
	    	return (long) Arrays.stream(a).mapToDouble(i->i*i).sum();
	    }
	    //use of filter
	    //Arrays.stream(a).filter(k->k==1).count();
	    //            array       variable->condition returning boolean  . lot fof functionsa re available
	    
	    static String[] generateBinSeq(int cnt)
	    {
	    	int loopvar=(int) Math.pow(2, cnt);
			String a[][] = new String[cnt+1][loopvar];
			
			for(int i=1;i<=cnt;i++)
			{
				int temp = (int) Math.pow(2, i);
				for(int j=0;j<temp;j++)
				{
					if(i==1 || i==2)
					{
						if(i==1)
						{
							a[i][0]="0";
							a[i][1]="1";
							continue;
						}
						else
						{
							a[2][0]="00";a[2][1]="01";a[2][2]="10";a[2][3]="11";
							continue;
						}
					}
					else
					{
						if(j<(temp/2))
						{
							a[i][j] = a[i-1][j]+"0"; 
						}
						else
						{
							a[i][j] = a[i-1][j%((temp+1)/2)]+"1";
						}
				
					}
						
				}
			}
	    
			return a[cnt];
	    }
	    
	 // USe this snippet for calling
//	    int ndx = algo.bs(a,key,0,a.length-1);
//		if(ndx<0)
//			ndx=0;
//		if(ndx==a.length)
//			ndx=a.length-1;

	    static public int bs(long a[],long key,int s,int e)
		{
			int m = (s+e)/2;
			if(s>e)
				return s;
				
			if(a[m]==key)
				return m;
			else if(a[m]>key)
				return bs(a,key,s,m-1);
			else
				return bs(a,key,m+1,e);
			
		}
	 // USe this snippet for calling
//	    int ndx = bsa(al,key,0,al.size()-1);
//		if(ndx<0)
//			ndx=0;
//		if(ndx==al.size())
//			ndx=al.size()-1;
	    static public int bsa(ArrayList<Long> a,long key,int s,int e)
		{
			int m = (s+e)/2;
			if(s>e)
				return s;
				
			if(a.get(m)==key)
				return m;
			else if(a.get(m)>key)
				return bsa(a,key,s,m-1);
			else
				return bsa(a,key,m+1,e);
			
		}
	    static public long fact(long a)
	    {
	    	long i,fact=1;  
	    	 long number=a;   
	    	  for(i=1;i<=number;i++){    
	    	      fact=fact*i;    
	    	  } 
	    	  return fact;
	    }
	    
	}
}
