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. 
import java.io.BufferedInputStream;
import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.Queue;
import java.util.Scanner;
import java.util.Stack;
import java.util.StringTokenizer;

public class Main {
	public static void main(String[] args) throws Exception {
		new Main();
	}
	
	InputReader ir;
	public Main()  throws Exception {
		ir = new InputReader(System.in);
		int tc = ir.nextInt();
		while(tc-->0)
		solve();
	}

	
	private void solve() {
		int n = ir.nextInt();
		int[] a = new int[n+2];
		for(int i=1; i<=n; i++) a[i] = ir.nextInt();
		a[0] = a[n+1] = 0;
		Stack<Integer> stack = new Stack();
		stack.add(0);
		int[] L = new int[n+1];
		for(int i=1; i<=n; i++) {
			while(!stack.isEmpty() && a[stack.peek()] >= a[i]) stack.pop();
			L[i] = stack.peek()+1;
			stack.add(i);
		}
		int[] R = new int[n+1];
		stack.clear();
		stack.add(n+1);
		for(int i=n; i>=1; i--) {
			while(!stack.isEmpty() && a[stack.peek()] >= a[i]) stack.pop();
			R[i] = stack.peek()-1;
			stack.add(i);
		}
		
		int[] arr = new int[n+1];
	//	for(int i=1; i<=n; i++) arr[i] = Integer.MAX_VALUE; 
		for(int i=1; i<=n; i++) {
			if(arr[a[i]] == 0) arr[a[i]] = R[i]-L[i]+1;
			else arr[a[i]] = Math.min(arr[a[i]], R[i]-L[i]+1);
		}
		for(int i=2; i<=n; i++) arr[i] = Math.min(arr[i], arr[i-1]);
	//	for(int i=1; i<=n; i++) System.out.println(arr[i]);
		StringBuilder ans = new StringBuilder();
		boolean isFalse = false;
		for(int i=1; i<=n; i++) {
			ans.append((arr[n-i+1]) >= i? "1":"0");
		}
		System.out.println(ans.toString());
	}
	
	
	class InputReader {
		private BufferedReader reader;
		private StringTokenizer tokenizer;
		
		public InputReader(InputStream is) {
			reader = new BufferedReader(new InputStreamReader(is), 8*2014);
			tokenizer = null;
		}
		
		public String next() {
			while(tokenizer == null || !tokenizer.hasMoreElements()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException();
				}
			}
			return tokenizer.nextToken();
		}
		
		public Integer nextInt() {
			return Integer.parseInt(next());
		}
	}
}
