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 Main
{
	public static void main(String[] args)throws Exception
	{
	    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	    PrintWriter pw=new PrintWriter(System.out);
	    int t=Integer.parseInt(br.readLine());
	    while(t-->0)
	    {
	        int n=Integer.parseInt(br.readLine());
	        String s[]=(br.readLine()).split(" ");
	        int p[]=new int[n+1];
	        for(int i=1;i<n+1;i++)
	        {
	            p[i]=Integer.parseInt(s[i-1]);
	        }
	        int ans[]=new int[n+1];
	        ans[1]=1;
	        ans[n]=1;
	       // char a='1';
	        //char b='0';
	        int l=1;
	        int r=n;
	        for(int i=n-1;i>=2;i--)
	        {
	            boolean L=false;
	            boolean R=false;//signifies if l & r are at the right position
	            while(l<r)
	            {
	                if(p[l]>i)
	                {
	                    l++;
	                }
	                else
	                {
	                    L=true;
	                }
	                if(p[r]>i)
	                {
	                    r--;
	                }
	                else
	                {
	                    R=true;
	                }
	                if(L & R)
	                {
	                    if((r-l+1)==i)
	                    {
	                        ans[i]=1;
	                    }
	                    else
	                    {
	                        ans[i]=0;
	                    }
	                    break;
	                }
	            }
	        }
	/*	if(n>1)
	        ans=a+ans;*/
	        for(int i=1;i<=n;i++)
	        pw.print(ans[i]);
	        pw.println();
	    }
	    pw.flush();
	}
}
