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.*;

/**
 * Created by Julia on 01.04.2017.
 */
public class SolverB {

    public static void main(String[] args) throws IOException {
        new SolverB().run();
    }

    StringTokenizer stok;
    BufferedReader br;
    PrintWriter pw;

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

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

    void run() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(new OutputStreamWriter(System.out));
//        br = new BufferedReader(new FileReader("input.txt"));
//        pw = new PrintWriter("output.txt");
        int k = nextInt();

        for(int test = 0; test < k; test++){
            int n = nextInt();
            int[] numToPos = new int[n+1];

            for(int i = 0; i < n; i++){
                int num = nextInt();
                numToPos[num] = i;
            }
            int[] segment = new int[n];
            segment[numToPos[1]] = 1;
            StringBuilder res = new StringBuilder();
            res.append('1');
            int min = numToPos[1];
            int max = numToPos[1];

            for(int i = 2; i < n+1; i++){
                segment[numToPos[i]] = 1;
                while(max < n && segment[max] == 1){
                    max++;
                }
                while(min >= 0 && segment[min] == 1){
                    min--;
                }
                if(max - min - 1 == i){
                    res.append('1');
                }else{
                    res.append('0');
                }
            }
            pw.println(res.toString());


        }
        pw.flush();
        pw.close();
    }

    class Num implements Comparable<Num>{
        int num, pos;

        public Num(int num, int pos) {
            this.num = num;
            this.pos = pos;
        }

        @Override
        public int compareTo(Num o) {
            return num - o.num;
        }
    }
}
