You are given a permutation p=[p_1, p_2, …, p_n] of integers from 1 to n. Let's call the number m (1 ≤ m ≤ n) beautiful, if there exists two indices l, r (1 ≤ l ≤ r ≤ n), such that the numbers [p_l, p_{l+1}, …, p_r] is a permutation of numbers 1, 2, …, m.

For example, let p = [4, 5, 1, 3, 2, 6]. In this case, the numbers 1, 3, 5, 6 are beautiful and 2, 4 are not. It is because:

  * if l = 3 and r = 3 we will have a permutation [1] for m = 1; 
  * if l = 3 and r = 5 we will have a permutation [1, 3, 2] for m = 3; 
  * if l = 1 and r = 5 we will have a permutation [4, 5, 1, 3, 2] for m = 5; 
  * if l = 1 and r = 6 we will have a permutation [4, 5, 1, 3, 2, 6] for m = 6; 
  * it is impossible to take some l and r, such that [p_l, p_{l+1}, …, p_r] is a permutation of numbers 1, 2, …, m for m = 2 and for m = 4. 



You are given a permutation p=[p_1, p_2, …, p_n]. For all m (1 ≤ m ≤ n) determine if it is a beautiful number or not.

Input

The first line contains the only integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. The next lines contain the description of test cases.

The first line of a test case contains a number n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the given permutation p. The next line contains n integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n, all p_i are different) — the given permutation p.

It is guaranteed, that the sum of n from all test cases in the input doesn't exceed 2 ⋅ 10^5.

Output

Print t lines — the answers to test cases in the order they are given in the input. 

The answer to a test case is the string of length n, there the i-th character is equal to 1 if i is a beautiful number and is equal to 0 if i is not a beautiful number.

Example

Input


3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2


Output


101011
11111
1001

Note

The first test case is described in the problem statement.

In the second test case all numbers from 1 to 5 are beautiful:

  * if l = 3 and r = 3 we will have a permutation [1] for m = 1; 
  * if l = 3 and r = 4 we will have a permutation [1, 2] for m = 2; 
  * if l = 2 and r = 4 we will have a permutation [3, 1, 2] for m = 3; 
  * if l = 2 and r = 5 we will have a permutation [3, 1, 2, 4] for m = 4; 
  * if l = 1 and r = 5 we will have a permutation [5, 3, 1, 2, 4] for m = 5. 


import java.util.Arrays;
import java.util.Scanner;

public class BEAUTIFULL_NUMBER {

	public static void main(String[] args) {

		Scanner nik = new Scanner(System.in);
		int t = nik.nextInt();
		StringBuilder st = new StringBuilder();
		while (t-- > 0) {
			int n = nik.nextInt();

			pair[] p = new pair[n];
			for (int i = 1; i <= n; i++) {
				p[i - 1] = new pair(nik.nextInt(), i);
			}

			Arrays.sort(p);
			st.append(1);
			int max = p[0].idx;
			int min = p[0].idx;

			for (int i = 1; i < n; i++) {
				int temp = Math.abs(p[i].idx - max);
				int temp2 = Math.abs(p[i].idx - min);
				int temp3 = Math.abs(max - min);
				max = max > p[i].idx ? max : p[i].idx;
				min = min > p[i].idx ? p[i].idx : min;
				if (temp >= p[i].d || temp2 >= p[i].d || temp3 >= p[i].d) {
					st.append(0);
				} else {
					st.append(1);
				}
			}
			st.append("\n");
		}
		System.out.println(st);
	}

	private static class pair implements Comparable<pair> {

		int d;
		int idx;

		pair(int d, int idx) {
			this.d = d;
			this.idx = idx;
		}

		public int compareTo(pair o) {
			return this.d - o.d;
		}
	}

}
