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.*;
import java.util.stream.Collectors;
import java.io.*;

public final class Main {
    public static void main(String[] args) {
        boolean debug = false;
        MyScanner in = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));
        if (debug) {
            in = new MyScanner(new File("./in"));
        }

        int tc = in.nextInt();
        while (tc-- > 0) {
            int n = in.nextInt();
            int[] arr = Arrays.asList(in.nextLine().split(" ")).stream().mapToInt(Integer::parseInt).toArray();

            int[] ans = new int[n + 1];
            Arrays.fill(ans, -1);
            int curr_mini = -1;

            for (int i = 0, j = n - 1; i <= j;) {
                if (ans[n] == -1) {
                    ans[n] = 1;
                } else {
                    int tmp;
                    // chose a new minimum
                    if (arr[i] > arr[j]) {
                        tmp = arr[i];
                        i++;
                    } else {
                        tmp = arr[j];
                        j--;
                    }

                    if (tmp < curr_mini || curr_mini == -1) {
                        curr_mini = tmp;
                    }

                    if (curr_mini > (j - i + 1)) {
                        ans[j - i + 1] = 1;
                    } else {
                        ans[j - i + 1] = 0;
                    }
                }
            }

            List<Integer> res = Arrays.stream(ans).boxed().collect(Collectors.toList());
            for (int i = 1; i < res.size(); i++) {
                out.print(res.get(i));
            }
            out.println();

        }

        out.close();

    }

    // -----------PrintWriter for faster output---------------------------------
    public static PrintWriter out;

    // -----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public MyScanner(File f) {
            try {
                br = new BufferedReader(new InputStreamReader(new FileInputStream(f)));
            } catch(FileNotFoundException e) {
                e.printStackTrace();
            }
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

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

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

    }
    // --------------------------------------------------------
}