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.*;
import java.util.*;
import java.math.*;
import java.lang.*;
 
import static java.lang.Math.*;
 
public class Solution implements Runnable 
{
	static class InputReader 
	{
		private InputStream stream;private byte[] buf = new byte[1024];
		private int curChar;
		private int numChars;
		private SpaceCharFilter filter;
		private BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        public InputReader(InputStream stream){this.stream = stream;}
		public int read(){if (numChars==-1) throw new InputMismatchException();if (curChar >= numChars){curChar = 0;try{numChars = stream.read(buf);}catch (IOException e){throw new InputMismatchException();	}if(numChars <= 0)return -1;}return buf[curChar++];}
     	public String nextLine(){String str = "";try{str = br.readLine();}catch (IOException e){e.printStackTrace();}return str;}
		public int nextInt(){int c = read();	while(isSpaceChar(c))c = read();	int sgn = 1; if (c == '-'){sgn = -1;c = read();	}int res = 0;do{if(c<'0'||c>'9') throw new InputMismatchException();res *= 10;res += c - '0';c = read();}while (!isSpaceChar(c));return res * sgn;}
        public long nextLong(){int c = read();while (isSpaceChar(c))c = read();int sgn = 1;if (c == '-'){sgn = -1;c = read();}long res = 0;do{if (c < '0' || c > '9')throw new InputMismatchException();res *= 10;res += c - '0';c = read();}while (!isSpaceChar(c));return res * sgn;}
		public double nextDouble(){int c = read();while (isSpaceChar(c))c = read();int sgn = 1;if (c == '-'){sgn = -1;c = read();}double res = 0;while (!isSpaceChar(c) && c != '.'){if (c == 'e' || c == 'E')return res * Math.pow(10, nextInt());if (c < '0' || c > '9')
		throw new InputMismatchException();res *= 10;res += c - '0';c = read();}if (c == '.'){c = read();double m = 1;while (!isSpaceChar(c)){if (c == 'e' || c == 'E')return res * Math.pow(10, nextInt());if (c < '0' || c > '9')throw new InputMismatchException();m /= 10;res += (c - '0') * m;c = read();}}return res * sgn;}
        public String readString(){int c = read();while (isSpaceChar(c))c = read();StringBuilder res = new StringBuilder();do{res.appendCodePoint(c);c = read();}while (!isSpaceChar(c));return res.toString();}
		public boolean isSpaceChar(int c){if (filter != null)return filter.isSpaceChar(c);return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;}
		public String next(){return readString();}
        public interface SpaceCharFilter{public boolean isSpaceChar(int ch);}
	}
		
	public static void main(String args[]) throws Exception{new Thread(null, new Solution(),"Main",1<<27).start();}

	public static long gcd(long a, long b) { 	if (a == 0) return b; return gcd(b % a, a);} 
  
	public static long findGCD(long arr[], int n) { 	long result = arr[0]; for (int i = 1; i < n; i++) result = gcd(arr[i], result);return result; }	 
	
	static void sortbycolomn(long arr[][], int col)
    { 
        Arrays.sort(arr, new Comparator<long[]>() { 
        @Override              
        public int compare(final long[] entry1,final long[] entry2) { 
            if (entry1[col] > entry2[col])return 1; 
            else return -1; 
        } 
        });   
    } 
    
    public void run()
    {
        InputReader in = new InputReader(System.in);
        PrintWriter w = new PrintWriter(System.out);
		
	    int t=in.nextInt();
	    while(t--!=0){
	        int n=in.nextInt();
	        int[] arr=new int[n+1];
	        for(int i=1;i<=n;i++)
	            arr[in.nextInt()]=i;
	        
	        int max=arr[1];
	        int min=arr[1];
	        char[] s=new char[n];
	        s[0]='1';
	        for(int i=2;i<=n;i++){
	            max=Math.max(max,arr[i]);
	            min=Math.min(min,arr[i]);
	            if(max-min+1==i)
	                s[i-1]='1';
	            else
	                s[i-1]='0';
	        }
	        w.println(new String(s));
	    }
	    
	    w.flush();
        w.close();
    }
}