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.StringTokenizer;

public class B {

    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int TC = sc.nextInt();
        for (int t = 0; t < TC; t++) {
            int n = sc.nextInt();
            int[] a = new int[n];
            int oneIdx = -1;
            for (int i = 0; i < n; i++) {
                a[i] = sc.nextInt();
                if (a[i] == 1) {
                    oneIdx = i;
                }
            }

            int l = oneIdx, r = oneIdx, max = 1;
            char[] ans = new char[n];
            ans[0] = '1';
            while (r - l + 1 < n) {
                if (l == 0 || r + 1 < n && a[l - 1] > a[r + 1]) {
                    r++;
                    max = Math.max(max, a[r]);
                }
                else {
                    l--;
                    max = Math.max(max, a[l]);
                }
                ans[r - l] = ((r - l + 1 == max) ? '1' : '0');
            }
            out.println(new String(ans));
        }
        out.flush();
        out.close();
    }

    static class Scanner
    {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream system) {br = new BufferedReader(new InputStreamReader(system));}
        public Scanner(String file) throws Exception {br = new BufferedReader(new FileReader(file));}
        public String next() throws IOException
        {
            while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }
        public String nextLine()throws IOException{return br.readLine();}
        public int nextInt() throws IOException {return Integer.parseInt(next());}
        public double nextDouble() throws IOException {return Double.parseDouble(next());}
        public char nextChar()throws IOException{return next().charAt(0);}
        public Long nextLong()throws IOException{return Long.parseLong(next());}
        public boolean ready() throws IOException{return br.ready();}
        public void waitForInput() throws InterruptedException {Thread.sleep(3000);}
    }
}
