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. 
// Working program with FastReader 
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.Scanner; 
import java.util.StringTokenizer;

// import jdk.javadoc.internal.doclets.toolkit.util.links.LinkFactory;

import java.util.*;
import java.io.*;
public class D_Rating_Compression{
    public static void main(String[] args) {
       FastReader sc=new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        int test = sc.nextInt();
        for(int t=0;t<test;t++){
            int n = sc.nextInt();
            int[] array = new int[n];
            HashMap<Integer,Integer> store = new HashMap<>();
            store.put(0,0);
            for(int i =0;i<n;i++){
                array[i] = sc.nextInt();
                store.put(array[i],store.getOrDefault(array[i],0)+1);
            }
            int left = 0;
            int right = n-1;
            int[] ans = new int[n];
            // if(array[left]==1 || array[right]==1)ans[n-1]=1;
            int find = 1;

            for(int i =n-1;i>=0;i--){
                if(store.containsKey(find) && store.get(find-1)<=0) {
                    ans[i]=1;
                    store.put(find,store.get(find)-1);
                    if(array[left]==find && array[right]==find){
                        break;
                    }
                    if(array[left]==find){
                        left++;
                    }
                    else if(array[right]==find){
                        right--;
                    }
                    else{
                        break;
                    }
                    find++;
                }
                else{
                    break;
                }
            }
            // check for along
            Arrays.sort(array);
            boolean yes = true;
            for(int i =0;i<n;i++){
                if(array[i]!=i+1){
                    yes = false;
                    break;
                }
            }
            if(yes)ans[0] = 1;
            
            for(int i =0;i<n;i++){
                out.print(ans[i]);
            }
            out.println();
        }
        out.close();
    }

    
    public static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 

        public FastReader() 
        { 
            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; 
        } 
}
}
