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.lang.*;
import java.awt.*;
import java.awt.geom.*;
import java.math.*;
import java.text.*;
import java.math.BigInteger.*;
import java.util.Arrays; 

public class   unwashme
{
  BufferedReader in;
  StringTokenizer as;
  int nums[];
  int l,r;
  boolean nums2[];

  public static void main (String[] args)
  {
    new unwashme  ();
  }
  
  
  
  
  public unwashme  ()
  {
    try
    {
      
      in = new BufferedReader (new InputStreamReader (System.in));
      int aa = nextInt();
      for(int z = 0;z<aa;z++)
      {
        int a = nextInt();
        nums = new int [a];
        nums2 = new boolean [a];
        boolean good[] = new boolean [a];
        good[0] = true;
        for(int x = 0;x<a;x++)
          nums[nextInt()-1] = x;
        l = nums[0];
        r = nums[0];
        nums2[nums[0]] = true;
        for(int x = 1;x<a;x++)
        {
          nums2[nums[x]] = true;
          while(l != 0 && nums2[l-1])
            l--;
          while(r != a-1 && nums2[r+1])
            r++;
          if(r -l == x)
            good[x] = true;
          
        }
        for(int x = 0;x<a;x++)
        {
          if(good[x])
            System.out.print(1);
          else
            System.out.print(0);
        }
        System.out.println();
      }


    }
    
    
    
    catch(IOException e)
    {
      
    }
  }
  
  
  
  
  
  
  
  String next () throws IOException
  {
    while (as == null || !as.hasMoreTokens ())
    {
      as = new StringTokenizer (in.readLine ().trim ());
    }
    
    
    return as.nextToken ();
  }
  
  
  
  long nextLong () throws IOException
  {
    return Long.parseLong (next ());
  }
  
  
  int nextInt () throws IOException
  {
    return Integer.parseInt (next ());
  }
  
  
  double nextDouble () throws IOException
  {
    return Double.parseDouble (next ());
  }
  
  
  String nextLine () throws IOException
  {
    return in.readLine ().trim ();
  }
}