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.*;


public class Main {

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(1, in, out);
        out.close();
    }

    static class Task {

        public void solve(int testNumber, InputReader input, PrintWriter out) {

            int t = input.nextInt();

            while(t-- > 0){
                int n = input.nextInt();

                int arr[] = new int[n];

                Map<Integer,Integer> map = new HashMap<>();
                for(int i = 0;i < n;i++){
                    arr[i] = input.nextInt();
                    map.put(arr[i],i);
                }

                Set<Integer> set = new HashSet<>();

                int left = map.get(1);
                int right = left;
                set.add(1);

                StringBuilder ans = new StringBuilder();
                ans.append('1');

                int num = 2;

                while(num <= n){

                    if(set.contains(num)){
                        num++;
                        if(set.size() == num - 1){
                            ans.append('1');
                        }
                        else{
                            ans.append('0');
                        }
//                        System.out.println("num:" + num + " set:" + set + " left:" + left + " right:" + right);
                        continue;
                    }

                    int idx = map.get(num);
//                    System.out.println("num: " + num + " idx:"+  idx);

                    int l = Math.abs(left - idx);
                    int r = Math.abs(right - idx);

                    if(l == r && l == 1){
                        if(idx > right){
                            right++;
                        }
                        else{
                            left--;
                        }
                        set.add(arr[idx]);
                        ans.append('1');
                    }
                    else if(l == 1){
                        if(set.size() + 1 == num){
                            ans.append('1');
                        }
                        else{
                            ans.append('0');
                        }
                        set.add(arr[idx]);
                        left--;
                    }
                    else if(r == 1){
                        if(set.size() + 1 == num){
                            ans.append('1');
                        }
                        else{
                            ans.append('0');
                        }
                        set.add(arr[idx]);
                        right++;
                    }
                    else{
                        ans.append('0');
                        int start,end;
                        if(l < r){
                            start = idx;
                            end = left;
                            left = idx;
                        }
                        else if(r < l){
                            start = right + 1;
                            end = idx + 1;
                            right = idx;
                        }
                        else{
                            if(idx < left){
                                start = idx;
                                end = left;
                                left = idx;
                            }
                            else{
                                start = left + 1;
                                end = idx + 1;
                                right = idx;
                            }
                        }

                        for(int i = start;i < end;i++){
                            set.add(arr[i]);
                        }
                    }

                    num++;
//                    System.out.println("num:" + num + " set:" + set + " left:" + left + " right:" + right);
                }



                out.println(ans);
                out.flush();
            }

        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

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

        public float nextFloat() {
            return Float.parseFloat(next());
        }

        public BigInteger nextBigInteger(){
            return new BigInteger(next());
        }


    }
}