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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.stream.IntStream;
import java.util.Scanner;
import java.util.OptionalInt;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author 4erneff
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        BBeautifulNumbers solver = new BBeautifulNumbers();
        solver.solve(1, in, out);
        out.close();
    }

    static class BBeautifulNumbers {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int t = in.nextInt();
            while (t-- > 0) {
                int n = in.nextInt();
                int[] arr = new int[n];
                for (int i = 0; i < n; ++i)
                    arr[i] = in.nextInt();
                handle(arr, n, out);
            }
        }

        private void handle(int[] arr, int sz, PrintWriter out) {
            char[] res = new char[sz];
            for (int i = 0; i < sz; ++i) {
                res[i] = '0';
            }

            int left = IntStream.range(0, arr.length)
                    .filter(i -> 1 == arr[i])
                    .findFirst()
                    .orElse(-1);
            int right = left;
            int maxEl = 1;
            res[0] = '1';

            while (right - left + 1 < sz) {
                int v1 = 10000000;
                int v2 = v1;

                if (left > 0) {
                    v1 = arr[left - 1];
                }
                if (right < sz - 1) {
                    v2 = arr[right + 1];
                }

                if (v1 < v2) {
                    left--;
                    maxEl = Math.max(maxEl, v1);
                } else {
                    right++;
                    maxEl = Math.max(maxEl, v2);
                }
                if (maxEl == (right - left + 1)) {
                    res[maxEl - 1] = '1';
                }
            }
            out.println(String.valueOf(res));
        }

    }
}

