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 implements Runnable {

    private static int MOD = 1_000_000_007;

    public static void main(String[] args) {
        // Run with 32MB stack
        Thread thread = new Thread(null, new Main(), "", 32 * 1024 * 1024);
        thread.start();
    }

    @Override
    public void run() {
        final Scanner scanner = new Scanner(System.in);
        solve(scanner);
    }

    static void solve(Scanner scanner) {
        int N = Integer.parseInt(scanner.nextLine());

        for (int n = 0; n < N; n++) {
            int size = Integer.parseInt(scanner.nextLine());
            int[] arr = Arrays.stream(scanner.nextLine().split(" ")).mapToInt(Integer::parseInt).toArray();

            if (size == 1) {
                System.out.println('1');
                continue;
            }

            TreeMap<Integer, Integer> set = new TreeMap<>();
            for (int i = 0; i < arr.length; i++) {
                set.put(arr[i], i);
            }

            Queue<Integer> added = new PriorityQueue<>(Comparator.comparingInt(x -> -x));

            StringBuilder ans = new StringBuilder();

            int left = set.get(1);
            int right = left;
            added.add(1);
            ans.append(1);

            TreeSet<Integer> rem = new TreeSet<>();
            for (int i = 2; i <= size; i++) {
                rem.add(i);
            }

            int target = set.get(2);
            for (int i = 2; i <= size; i++) {
                if (target > right) {
                    added.add(arr[++right]);
                    rem.remove(arr[right]);
                    if (target == right && !rem.isEmpty()) {
                        target = set.get(rem.first());
                    }
                    if (added.peek() == i) {
                        ans.append(1);
                    } else {
                        ans.append(0);
                    }
                } else {
                    added.add(arr[--left]);
                    rem.remove(arr[left]);
                    if (target == left && !rem.isEmpty()) {
                        target = set.get(rem.first());
                    }
                    if (added.peek() == i) {
                        ans.append(1);
                    } else {
                        ans.append(0);
                    }
                }
            }

            System.out.println(ans.toString());
        }
    }

}
