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 static java.lang.System.exit;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Stack;
import java.util.StringTokenizer;


public class Task4 {
	//correct but slow, faster in task5;
	static void solve() throws IOException{
		int n = scanInt();
		int[] arr = new int[n];
		for (int i = 0; i < n; i++) {
			arr[i] = scanInt();
		}
		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < n; i++) {
			sb.append('1');
		}
		
		//first min to the left;
		int[] left = new int[n];
		left[0] = -1;
		Stack<Integer> st = new Stack<Integer>();
		st.push(0);
		for (int i = 1; i < n; i++) {
			while(!st.isEmpty() && arr[st.peek()] > arr[i]) {
				st.pop();
			}
			if (!st.isEmpty()) {
				left[i] = st.peek();
			}else {
				left[i] = -1;
			}
			st.push(i);
		}
		//first min to the right;
		int[] right = new int[n];
		right[n-1] = n;
		st = new Stack<Integer>();
		st.push(n-1);
		for (int i = n-2; i >= 0; i--) {
			while(!st.isEmpty() && arr[st.peek()] >= arr[i]) {
				st.pop();
			}
			if (!st.isEmpty()) {
				right[i] = st.peek();
			}else {
				right[i] = n;
			}
			st.push(i);
		}
		int[][] info = new int[n+1][2];
		int start = 1;
		for (int i = 0; i < n; i++) {
			int distLeft = i - left[i] - 1;
			int distRight = right[i] - i - 1;
			int dist = distLeft + distRight;
			if (info[arr[i]][0] > 0) {
				sb.setCharAt(0, '0');
				if (dist == info[arr[i]][1])  sb.setCharAt(dist, '0');
				dist = Math.max(dist, info[arr[i]][1]);
				distLeft = distRight = dist;
			}
			if (Math.min(distLeft, distRight) > 0) {
				for (int j = start; j < dist; j++) {
					sb.setCharAt(j, '0');
				}
				start = Math.max(start, dist - 1);
			}
			info[arr[i]][0] = 1;
			info[arr[i]][1] = dist;
		}
		for (int i = 1; i < n+1; i++) {
			if (info[i][0] == 0) {
				for (int j = 0; j < n - i + 1; j++) {
					sb.setCharAt(j, '0');
				}
				break;
			}
		}
		String answ = sb.toString();
		
		out.println(answ);
	}
	
	static int scanInt() throws IOException{
		return Integer.parseInt(scanString());
	}
	
	static long scanLong() throws IOException{
		return Long.parseLong(scanString());
	}
	
	static String scanString() throws IOException{
		while(tok == null || !tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine());
		}
		return tok.nextToken();
	}
	
	static void printCase() {
		out.print("Case #" + test + ": ");
	}

	static void printlnCase() {
		out.println("Case #" + test + ":");
	}
	
	static BufferedReader in;
	static PrintWriter out;
	static StringTokenizer tok; 
	static int test;
	
	public static void main(String[] args) throws IOException{
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			int tests = scanInt();
			for(test = 1; test <= tests; test++) {
				solve();
			}
			in.close();
			out.close();
		}catch(Throwable e) {
			e.printStackTrace();
			exit(1);
		}
	}
}