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.*;
import java.math.*;
 
public class beatu2{

    public static PrintWriter out;
      
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;
 
        public MyScanner(){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){
        MyScanner in=new MyScanner();
        out=new PrintWriter(new BufferedOutputStream(System.out));
        
        int t=in.nextInt();
        while(t-->0){
            int n=in.nextInt(), posOf[]=new int[n+1], ans[]=new int[n+1];
            
            for(int i=0; i<n; i++){
                posOf[in.nextInt()]=i;
            }
            int l=n, r=0;
            for(int i=1; i<=n; i++){
                l=Math.min(l, posOf[i]);
                r=Math.max(r, posOf[i]);
                if(r-l+1==i)
                    ans[i]=1;
            }
            for(int i=1; i<=n; i++){
                if(ans[i]==1)
                    out.printf("1");
                else 
                    out.printf("0");
            }
            out.printf("\n");
        }




        out.close();
    }      

    
}