On the competitive programming platform CodeCook, every person has a rating graph described by an array of integers a of length n. You are now updating the infrastructure, so you've created a program to compress these graphs.

The program works as follows. Given an integer parameter k, the program takes the minimum of each contiguous subarray of length k in a.

More formally, for an array a of length n and an integer k, define the k-compression array of a as an array b of length n-k+1, such that $$$b_j =min_{j≤ i≤ j+k-1}a_i$$$

For example, the 3-compression array of [1, 3, 4, 5, 2] is [min\{1, 3, 4\}, min\{3, 4, 5\}, min\{4, 5, 2\}]=[1, 3, 2].

A permutation of length m is an array consisting of m distinct integers from 1 to m in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (m=3 but there is 4 in the array).

A k-compression array will make CodeCook users happy if it will be a permutation. Given an array a, determine for all 1≤ k≤ n if CodeCook users will be happy after a k-compression of this array or not.

Input

The first line contains a single integer t (1≤ t≤ 10^4) — the number of test cases.

The first line of the description of each test case contains a single integer n (1≤ n≤ 3⋅ 10^5) — the length of the array.

The second line of the description of each test case contains n integers a_1,…,a_n (1≤ a_i≤ n) — the elements of the array.

It is guaranteed, that the sum of n for all test cases does not exceed 3⋅ 10^5.

Output

For each test case, print a binary string of length n. 

The k-th character of the string should be 1 if CodeCook users will be happy after a k-compression of the array a, and 0 otherwise. 

Example

Input


5
5
1 5 3 4 2
4
1 3 2 1
5
1 3 3 3 2
10
1 2 3 4 5 6 7 8 9 10
3
3 3 2


Output


10111
0001
00111
1111111111
000

Note

In the first test case, a=[1, 5, 3, 4, 2].

  * The 1-compression of a is [1, 5, 3, 4, 2] and it is a permutation. 
  * The 2-compression of a is [1, 3, 3, 2] and it is not a permutation, since 3 appears twice. 
  * The 3-compression of a is [1, 3, 2] and it is a permutation. 
  * The 4-compression of a is [1, 2] and it is a permutation. 
  * The 5-compression of a is [1] and it is a permutation. 
//make sure to make new file!
import java.io.*;
import java.util.*;

public class DG12{
   
   public static void main(String[] args)throws IOException{
      BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter out = new PrintWriter(System.out);
      
      int t = Integer.parseInt(f.readLine());
      
      for(int q = 1; q <= t; q++){

         int n = Integer.parseInt(f.readLine());
      
         StringTokenizer st = new StringTokenizer(f.readLine());
         
         int[] array = new int[n];
         int[] freq = new int[n+1];
         for(int k = 0; k < n; k++){
            array[k] = Integer.parseInt(st.nextToken());
            freq[array[k]]++;
         }
         
         int[] indexof = new int[n+1];
         Arrays.fill(indexof,-1);
         int num1 = 0;
         for(int k = 0; k < n; k++){
            if(freq[array[k]] == 1){
               indexof[array[k]] = k;
               num1++;
            }
         }
         
         int[] answer = new int[n];
         
         int l = 0;
         int r = n-1;
         
         for(int k = 1; k < n; k++){
            if(freq[k] > 0 && (k==1 || (freq[k-1] == 1 && (indexof[k-1] == l|| indexof[k-1] == r)))){
               answer[n-k] = 1;
               if(indexof[k-1] == l) l++;
               else if(indexof[k-1] == r) r--;
            } else {
               break;
            }
         }
         
         if(num1 == n) answer[0] = 1;
         
         StringJoiner sj = new StringJoiner("");
         for(int k = 0; k < n; k++){
            sj.add("" + answer[k]);
         }
         out.println(sj.toString());
      

      }
      
      
      
      
      out.close();
   }
   
      
}