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.lang.*;
import java.lang.reflect.Array;
import java.io.*;
import java.math.*;
import java.text.DecimalFormat;
public class Prac{     
    static class InputReader { 
        private final InputStream stream;
	private final byte[] buf = new byte[8192];
	private int curChar, snumChars;
 	public InputReader(InputStream st) {
            this.stream = st;
	} 
	public int read() {
            if (snumChars == -1)
	 	throw new InputMismatchException();
            if (curChar >= snumChars) {
		curChar = 0;
                try {
                    snumChars = stream.read(buf);
		} 
                catch (IOException e) {
                    throw new InputMismatchException();
           	}
		if (snumChars <= 0)
                    return -1;
            }
            return buf[curChar++];
	}
        public int ni() {
            int c = read();
            while (isSpaceChar(c)) {
               	c = read();
            }
            int sgn = 1;
            if (c == '-') {
		sgn = -1;
		c = read();
            }
            int res = 0;
            do {
		res *= 10;
		res += c - '0';
		c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
	} 
	public long nl() {
            int c = read();
            while (isSpaceChar(c)) {
            c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            long res = 0;
            do {
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
	} 
	public int[] nia(int n) {
            int a[] = new int[n];
            for (int i = 0; i < n; i++) {
		a[i] = ni();
            }
            return a;
	} 
	public String rs() {
            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 String nextLine() {
            int c = read();
            while (isSpaceChar(c))
		c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            } while (!isEndOfLine(c));
            return res.toString();
        } 
	public boolean isSpaceChar(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
	}
 
	private boolean isEndOfLine(int c) {
            return c == '\n' || c == '\r' || c == -1;
	} 
    }
    static PrintWriter w = new PrintWriter(System.out);
    
  
    
    public static void main(String[] args) throws IOException {   
        InputReader sc=new InputReader(System.in);
        int t=sc.ni();
        while(t-->0){
            int in=-1;
           int n=sc.ni();
           int arr[]=new int[n+1];
           for(int i=1;i<=n;i++){
               arr[i]=sc.ni();
               if(arr[i]==1){
                   in=i;
               }
           }
           int l=in,r=in;
           int ans[]=new int[n+1];
           ans[1]=1;
           
           for(int i=2;i<=n;i++){
               while(r<=n&&arr[r]<=i){
                   r++;
               }
               while(l>0&&arr[l]<=i){
                   l--;
               }
               if((r-l-1)==i){
                   ans[i]=1;
               }
               else{
                   ans[i]=0;
               }
           }
           for(int i=1;i<=n;i++){
               w.print(ans[i]);
           }
           w.println();
        }
           

        w.close();
    }
}
