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.*;

public class Q3 {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        long pre[] = new long[200009];
        for (int i = 1; i < pre.length; i++) {
            pre[i] += pre[i - 1] + i;
        }
        int t = in.nextInt();
        while (t-- > 0) {
            int N = in.nextInt(), l = 0, r = 0;
            int arr[] = new int[N];
            for (int i = 0; i < N; i++) {
                arr[i] = in.nextInt();
                if (arr[i] == 1)
                    l = r = i;
            }

            StringBuilder str = new StringBuilder();
            str.append('1');
            long sum = 1;
            for (int i = 1; i < N; i++) {

                if (l != 0 && r != N - 1) {
                    int a = arr[l - 1], b = arr[r + 1];
                    if (a < b) {
                        l--;
                        sum += a;
                    } else {
                        r++;
                        sum += b;
                    }
                } else if (l != 0) {
                    sum += arr[l - 1];
                    l--;
                } else {
                    sum += arr[r + 1];
                    r++;
                }

                if (sum == pre[i + 1])
                    str.append('1');
                else
                    str.append('0');
            }
            out.println(str);
        }
        out.close();
    }


    ///          PREFIX AND SUFFIX REMOVAL D PROBLEM
//    public static void main(String[] args) {
//        Scanner in = new Scanner(System.in);
//        PrintWriter out = new PrintWriter(System.out);
//
//        int N = in.nextInt(), M = in.nextInt(), j = 0;
//        int arr[] = new int[N];
//        HashMap<String, Integer> map = new HashMap<>();
//
//        while (j<M) {
//            for (int i = 0; i < N; i++)
//                arr[i] = in.nextInt();
//            StringBuilder t = new StringBuilder();
//            for (int i = 0; i < N; i++) {
//                t=new StringBuilder();
//                for (int k = i; k < N; k++) {
//                    t.append(arr[k]); t.append("^");
//                    map.put(t.toString(), map.getOrDefault(t.toString(), 0) + 1);
//                }
//            }
//            j++;
//        }
//
//        long ans = 0;
//        for (String key : map.keySet()) {
//            if (map.get(key) == M)
//                ans++;
//        }
//        out.println(ans);
//
//        out.close();
//    }
}