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. 

// Problem: D. Rating Compression
// Contest: Codeforces - Codeforces Global Round 12
// URL: https://codeforces.com/contest/1450/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

import java.io.*;
import java.util.*;

public class Main {
    private static PrintWriter pw = new PrintWriter(System.out);
    private static InputReader sc = new InputReader();
    private static final int intmax = Integer.MAX_VALUE, intmin = Integer.MIN_VALUE;

    static class Pair{
        int first, second;
        Pair(int first, int second){
            this.first = first;
            this.second = second;
        }
    }

    static class InputReader{
        private static BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
        private static StringTokenizer tk;

        private void next()throws IOException{
            while(tk == null || !tk.hasMoreTokens())   
                tk = new StringTokenizer(r.readLine());
        }

        private int nextInt()throws IOException{
            next();
            return Integer.parseInt(tk.nextToken());
        }

        private int[] intArray(int n)throws IOException{
            next();
            int arr[] = new int[n];

            for(int i=0; i<n; i++)
                arr[i] = nextInt();

            return arr;
        }
    }

    public static void main(String args[])throws IOException{
        int t = sc.nextInt();
        while(t-->0)    solve();
        pw.flush();
        pw.close();
    }

    private static void solve()throws IOException{
		int n = sc.nextInt(), a[] = sc.intArray(n), l[] = new int[n], r[] = new int[n], max[] = new int[n + 1];
		
		Deque<Integer> deque = new ArrayDeque<>();
		
		for(int i = 0; i < n; i++){
			while(!deque.isEmpty() && a[deque.peekLast()] >= a[i])
				deque.pollLast();
			
			if(deque.isEmpty())
				l[i] = -1;
			else
				l[i] = deque.peekLast();
				
			deque.addLast(i);
		}
		
		deque.clear();
		
		for(int i = n - 1; i >= 0; i--){
			while(!deque.isEmpty() && a[deque.peekLast()] >= a[i])
				deque.pollLast();
			
			if(deque.isEmpty())
				r[i] = n;
			else
				r[i] = deque.peekLast();
				
			deque.addLast(i);
		}
		
		for(int i = 0; i < n; i++)
			max[a[i]] = Math.max(max[a[i]], r[i] - l[i] - 1);
		
		for(int i = 2; i <= n; i++)
			max[i] = Math.min(max[i], max[i - 1]);
			
		// pw.println(Arrays.toString(max));
			
		for(int i = n; i >= 1; i--){
			if(max[i] == n - i + 1)
				pw.print(1);
			else
				pw.print(0);
		}
		
		pw.println();
    }
}