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.util.Arrays;
import java.util.StringTokenizer;

public class barray {
	public static void main(String[] args) {
		FS s=new FS();
		int nq=s.nextInt();
		while(nq-->0) {
			int n=s.nextInt();
			int[] arr=new int[n];
			int[] idxOf=new int[n];
			for(int i=0;i<n;i++) {
				arr[i]=s.nextInt()-1;
				idxOf[arr[i]]=i;
			}
			int lpr=idxOf[0];
			int rpr=idxOf[0];
			for(int i=0;i<n;i++) {
				while(lpr>0&&arr[lpr-1]<=i) {
					lpr--;
				}
				while(rpr<n-1&&arr[rpr+1]<=i)rpr++;
				if(rpr-lpr+1==i+1)System.out.print(1);
				else System.out.print(0);
			}
			System.out.println();
		}
	}
	static class dsu{
		int[] s,size;
		int comps;
		public dsu(int n) {
			Arrays.fill(s=new int[comps=n],-1);
			Arrays.fill(size=new int[n], 1);
		}
		public int find(int i) {
			return s[i]<0?i:(s[i]=find(s[i]));
		}
		public boolean union(int a, int b) {
			if((a=find(a))==(b=find(b)))return false;
			if(s[a]==s[b])s[a]--;
			if(s[a]<=s[b])s[b]=a;else s[a]=b;
			size[a]=size[b]+=size[a];
			return comps-->0;
		}
	}
	static class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while(!st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch(Exception e) {}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
}
