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.math.*;
import java.util.*;


// @author : Dinosparton 

public class test {
	 
	   static class Pair{ 
		   long x;
		   long y;
		   
		   Pair(long x,long y){ 
			   this.x = x;
			   this.y = y;
			   
		   }
	   }
	  
	   static class Compare { 
		   
		     void compare(Pair arr[], int n) 
		    { 
		        // Comparator to sort the pair according to second element 
		        Arrays.sort(arr, new Comparator<Pair>() { 
		            @Override public int compare(Pair p1, Pair p2) 
		            { 
		                return (int)(p1.x - p2.x); 
		            } 
		        }); 
		  
//		        for (int i = 0; i < n; i++) { 
//		            System.out.print(arr[i].x + " " + arr[i].y + " "); 
//		        } 
//		        System.out.println(); 
		    } 
		} 
	 
	   static class Scanner {
	        BufferedReader br;
	        StringTokenizer st;
	 
	        public Scanner()
	        {
	            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;
	        }
	    }
	   
	   public static void main(String args[]) throws Exception { 
		
		   Scanner sc = new Scanner();
		   StringBuffer ans = new StringBuffer();
		   
		   int tc = sc.nextInt();
		   
		   while(tc-->0) { 
			   
			 int n = sc.nextInt();
			 
			 int freq[] = new int[200001];
			 
			 int a[] = new int[n];
			 
			 for(int i=0;i<n;i++) { 
				 a[i] = sc.nextInt();
				 
				 freq[a[i]] = i+1;
			 }
			 
			 int min = Integer.MAX_VALUE;
			 int max = Integer.MIN_VALUE;
			 
			 for(int i=1;i<=n;i++) { 
				 
				 int x = freq[i];
				 
				 if(x<min)
					 min = x;
				 if(x>max)
					 max = x;
				 
				 if(max-min+1==i)
					 ans.append(1);
				 else
					 ans.append(0);
			 }
			 
			 ans.append("\n");
		   }
			System.out.println(ans);
	   }
}  




