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.PrintWriter;
import java.util.Scanner;

public class BeautifulNumbersB extends PrintWriter {
    //this trick improves performances
    BeautifulNumbersB() {
        super(System.out);
    }

    public static void main(String[] $) {
        BeautifulNumbersB o = new BeautifulNumbersB();
        o.main();
        o.flush();
    }

    void main() {
        Scanner sc = new Scanner(System.in);
        int count = sc.nextInt();
        //use this if just a single test
        //int count = 1;
        while (count-- > 0) {
            final int n = sc.nextInt();
            final int[] numbers = new int[n + 1];
            final StringBuilder result = new StringBuilder("1");
            //I store the position for the value
            for (int i = 1; i <= n; i++) {
                numbers[sc.nextInt()] = i;
            }
            if(n==1){
                println("1");
                continue;
            }
            int max = numbers[1];
            int min = max;
            long sum = max;

            for (int j = 2; j < n; j++) {
                min = Math.min(min, numbers[j]);
                max = Math.max(max, numbers[j]);
                sum += numbers[j];
                //comparing theoretical sum with real sum
                final long sumT = (long) max * (max + 1) / 2 - (long) (min - 1) * min / 2;
                if (sum == sumT) {
                    result.append("1");
                } else {
                    result.append("0");
                }
            }
            result.append("1");
            println(result);
        }
        sc.close();
    }
}
