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.BufferedReader;
import java.io.BufferedWriter;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) throws IOException {
        FastReader fr = new FastReader();
        FastWriter fw = new FastWriter();
        int t = fr.nextInt();

        while (t-- > 0) {
            int n = fr.nextInt();
            int arr[] = new int[n + 1];
            for (int i = 1; i <= n; i++) {
                arr[fr.nextInt()] = i;
            }
            int min = arr[1];
            int max = arr[1];
            StringBuilder sb = new StringBuilder("1");
            for (int i = 2; i <= n; i++) {
                min = Math.min(min, arr[i]);
                max = Math.max(max, arr[i]);
                if (max - min + 1 == i) sb.append("1");
                else sb.append("0");
            }
            fw.println(sb);
        }
    }

}


class FastReader {
    public FastReader() {
    }

    public FastReader(String fileName) throws FileNotFoundException {
        this.br = new BufferedReader(new FileReader(fileName));
    }

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st;

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

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

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

    double nextDouble() throws IOException {
        return Double.parseDouble(nextString());
    }
}

class FastWriter {
    public FastWriter() {
    }

    public FastWriter(String fileName) throws IOException {
        this.bw = new BufferedWriter(new FileWriter(fileName));
    }

    BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));

    void print(Object o) throws IOException {
        bw.write(o.toString());
        bw.flush();
    }

    void println(Object o) throws IOException {
        print(o.toString() + "\n");
    }

    void printSp(Object o) throws IOException {
        print(o.toString() + " ");
    }
}