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.*;
public class Main {
	static class NODE
	{
		public int date;
		public int index;
		NODE(){};
		NODE(int x,int y)
		{
			date=x;index=y;
		}
	}
	final static int maxn=(int)2e5+10;
	static NODE a[]=new NODE[maxn];
	static class cmp implements Comparator <NODE>
	{
		public int compare(NODE a,NODE b)
		{
			if(a.date<b.date)return -1;//从小到大排序为-1，从大到小排序为1
			else if(a.date==b.date)return 0;
			else
				return 1;
		}		 
	}
	static int vis[]=new int [maxn];
public static void main(String args[])
{
	//System.out.println(1);
	Scanner input=new Scanner(System.in);
	for(int i=1;i<=maxn-5;i++)
	{
		a[i]=new NODE();
	}
	int t=input.nextInt();
	while(t-->0)
	{
		int n=input.nextInt();
		for(int i=1;i<=n;i++)vis[i]=0;
		for(int i=1;i<=n;i++)
			{a[i].date=input.nextInt();
			a[i].index=i;
			}
		Arrays.sort(a,1,1+n,new cmp());
		int left,right;
		vis[1]=1;
		left=a[1].index;right=a[1].index;
		int num=1;
		for(int i=2;i<=n;i++)
		{
			num++;
			if(a[i].index<left)left=a[i].index;
			if(a[i].index>right)right=a[i].index;
			if(right-left+1==num)vis[i]=1;
		}
		for(int i=1;i<=n;i++)
			System.out.print(vis[i]);
		System.out.println();
			}
		
	}
}

	 	 	 	 	    	  			    		 	 	