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.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class b {

	public static void main(String[] args) throws Exception {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		sc = new FastScanner(in);
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			int[] arr = new int[n];
			for(int i = 0 ;i < n;i++)arr[i] =sc.nextInt();
			int lo , hi , ind = -1;
			for(int i = 0 ;i < n;i++) if(arr[i]==1)ind = i;
			lo  = ind;
			hi = ind;
			int[] ans = new int [n];
			for(int j = 1; j <= n;j++) {
				while(lo >0 && arr[lo-1] <= j ) {
					lo--;
				}
				while(hi < n -1&& arr[hi+1] <= j ) {
					hi++;
				}		
				
				if( hi - lo +1 == j)ans[j-1]=1;
			}
			for(int val : ans) {
				out.print(val);
			}
			out.println();
		}
		out.flush();
	}

	static BufferedReader in;
	static FastScanner sc;
	static PrintWriter out;

	static class FastScanner {
		BufferedReader in;
		StringTokenizer st;

		public FastScanner(BufferedReader in) {
			this.in = in;
		}

		public String nextToken() throws Exception {
			while (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(in.readLine());
			}
			return st.nextToken();
		}

		public int nextInt() throws Exception {
			return Integer.parseInt(nextToken());
		}

		public long nextLong() throws Exception {
			return Long.parseLong(nextToken());
		}

		public double nextDouble() throws Exception {
			return Double.parseDouble(nextToken());
		}
	}
}
