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.math.BigInteger;
import java.util.*;

import static java.lang.Math.*;
import static java.util.Arrays.*;

public class Main {

    public static void main(String[] args) throws IOException {
        FastScanner in = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int t = in.nextInt();
        while (t-- > 0) {
            int n = in.nextInt();
            int a[] = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = in.nextInt();
            }
            Stack<Integer> st = new Stack<>();
            int gl[] = new int[n], gr[] = new int[n];
            fill(gl, -1);
            fill(gr, n);
            for (int i = 0; i < n; i++) {
                while (!st.isEmpty() && a[st.peek()] < a[i]) {
                    gr[st.pop()] = i;
                }
                st.push(i);
            }
            st.clear();
            for (int i = n - 1; i >= 0; i--) {
                while (!st.isEmpty() && a[st.peek()] < a[i]) gl[st.pop()] = i;
                st.push(i);
            }
            int ans[] = new int[n];
            for (int i = 0; i < n; i++) {
                if (gr[i] - gl[i] - 1 == a[i]) ans[a[i] - 1] = 1;
            }
            for (int i = 0; i < n; i++) {
                out.print(ans[i]);
            }
            out.println();
        }
        out.close();
    }
}

class FastScanner {
    BufferedReader br;
    StringTokenizer st;

    FastScanner(File f) throws FileNotFoundException {
        br = new BufferedReader(new FileReader(f));
    }

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

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

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

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