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.*;
import java.io.*;
public class Q1Div3 
{
	public static void main(String[] $) throws Exception
	{
		Q1Div3 q1Div3=new Q1Div3();
		q1Div3.main();
	}
	void main() throws Exception
	{
		Scan scan = new Scan();
		Print print = new Print();
		int T = scan.scanInt();
		while (T-- > 0)
		{
			int N=scan.scanInt();
			int arr[]=new int[N];
			int pos[]=new int[N];
			for (int i = 0; i < N; i++) 
			{
				arr[i] = scan.scanInt();
				pos[arr[i]-1]=i;
			}
			StringBuilder ans=new StringBuilder();
			int max=-1;
			int min=1000000;
			for(int i=0;i<N;i++)
			{
				max=Math.max(max,pos[i]);
				min=Math.min(min,pos[i]);
				if(max-min==i)
				{
					ans.append("1");
				}
				else
				{
					ans.append("0");
				}
			}
			print.println(ans.toString());
		}
		print.close();
	}
	static class Print {
		private final BufferedWriter bw;

		public Print() {
			this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
		}

		public void print(Object object) throws IOException {
			bw.append("" + object);
		}

		public void println(Object object) throws IOException {
			print(object);
			bw.append("\n");
		}

		public void close() throws IOException {
			bw.close();
		}
	}

	static class Scan {
		private byte[] buf = new byte[1024 * 1024];
		private int index;
		private InputStream in;
		private int total;

		public Scan() {
			in = System.in;
		}

		public int scan() throws IOException {
			if (total < 0)
				throw new InputMismatchException();
			if (index >= total) {
				index = 0;
				total = in.read(buf);
				if (total <= 0)
					return -1;
			}
			return buf[index++];
		}

		public int scanInt() throws IOException {
			int integer = 0;
			int n = scan();
			while (isWhiteSpace(n))
				n = scan();
			int neg = 1;
			if (n == '-') {
				neg = -1;
				n = scan();
			}
			while (!isWhiteSpace(n)) {
				if (n >= '0' && n <= '9') {
					integer *= 10;
					integer += n - '0';
					n = scan();
				} else
					throw new InputMismatchException();
			}
			return neg * integer;
		}
		private boolean isWhiteSpace(int n) {
			if (n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1)
				return true;
			return false;
		}
	}
}
