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.*;
import java.util.*;

public class Main {


    public static void main(String[] args) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        int q = nextInt();
        while (q-- > 0) {
            int n = nextInt();
            point a[] = new point[n];
            long pref[] = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = new point(nextInt(), i);
            }
            boolean ans[] = new boolean[n];
            Arrays.sort(a, new point());
            ans[0] = true;
            int minx = a[0].ind;
            int maxx = a[0].ind;
            for (int i = 1; i < n; i++) {
                if (ans[i - 1]) {
                    int ind = a[i].ind;
                    if (minx - 1 == ind || maxx + 1 == ind) ans[i] = true;
                } else {
                    int ind = a[i].ind;
                    if (maxx - minx == i) {
                        if (minx < ind && ind < maxx) ans[i] = true;
                    }
                }
                minx = Math.min(a[i].ind, minx);
                maxx = Math.max(a[i].ind, maxx);
            }
            for (int i = 0; i < n; i++) {
                if (ans[i]) out.print("1");
                else out.print("0");
            }
            out.println();
        }
        out.close();
    }

    static BufferedReader br;
    static StringTokenizer st = new StringTokenizer("");

    static String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(br.readLine());
        }
        return st.nextToken();
    }

    static int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    static long nextLong() throws IOException {
        return Long.parseLong(next());
    }

}

class point implements Comparator<point> {
    int v, ind;

    public point(int v, int ind) {
        this.v = v;
        this.ind = ind;
    }

    public point() {
    }

    @Override
    public int compare(point o1, point o2) {
        return Integer.compare(o1.v, o2.v);
    }
}