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 {

    static FastReader in;
    static PrintWriter out;
    static Random rand = new Random();

    static void solve() {
        int n = in.nextInt();

        int[] p = new int[n + 1];
        int[] pos = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            p[i] = in.nextInt();
            pos[p[i]] = i;
        }

        int[] ans = new int[n + 1];
        int[] l = new int[n + 1];
        int[] r = new int[n + 1];

        Stack<Integer> st = new Stack<>();
        for (int i = 1; i <= n; i++) {
            while (!st.empty() && st.peek() < p[i]) {
                st.pop();
            }
            if (st.empty()) {
                l[p[i]] = 1;
            } else {
                l[p[i]] = pos[st.peek()] + 1;
            }
            st.push(p[i]);
        }
        st.clear();
        for (int i = n; i >= 1; i--) {
            while (!st.empty() && st.peek() < p[i]) {
                st.pop();
            }
            if (st.empty()) {
                r[p[i]] = n;
            } else {
                r[p[i]] = pos[st.peek()] - 1;
            }
            st.push(p[i]);
        }

        for (int i = 1; i <= n; i++) {
            if (r[i] - l[i] + 1 == i) {
                out.print(1);
            } else {
                out.print(0);
            }
        }
        out.println();
    }

    public static void main(String[] args) {
        in = new FastReader(System.in);
//        in = new FastReader(new FileInputStream("input.txt"));
        out = new PrintWriter(System.out);
//        out = new PrintWriter(new FileOutputStream("output.txt"));


        int q = in.nextInt();

        while (q-- > 0) {
            solve();
        }

        out.close();
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        FastReader(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        Integer nextInt() {
            return Integer.parseInt(next());
        }

        Long nextLong() {
            return Long.parseLong(next());
        }

        Double nextDouble() {
            return Double.parseDouble(next());
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(nextLine());
            }
            return st.nextToken();
        }

        String nextLine() {
            String s = "";
            try {
                s = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return s;
        }
    }
}