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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.util.Comparator;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        inputClass in = new inputClass(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        BBeautifulNumbers solver = new BBeautifulNumbers();
        solver.solve(1, in, out);
        out.close();
    }

    static class BBeautifulNumbers {
        public void solve(int testNumber, inputClass sc, PrintWriter out) {
            int t = sc.nextInt();
            while (t > 0) {
                t--;
                int n = sc.nextInt();
                BBeautifulNumbers.Pair[] tab = new BBeautifulNumbers.Pair[n];
                for (int i = 0; i < n; i++) {
                    tab[i] = new BBeautifulNumbers.Pair(sc.nextInt(), i);
                }
                Arrays.sort(tab, Comparator.comparingInt(a -> a.x));
                StringBuilder ans = new StringBuilder();
                ans.append(1);
                int maxl = tab[0].y;
                int maxr = tab[0].y;
                for (int i = 1; i < n; i++) {
                    maxl = Math.min(tab[i].y, maxl);
                    maxr = Math.max(tab[i].y, maxr);
                    if (maxr - maxl + 1 == i + 1) {
                        ans.append(1);
                    } else {
                        ans.append(0);
                    }
                }
                out.println(ans);
            }
        }

        static class Pair {
            int x;
            int y;
            int z;

            public Pair() {
            }

            public Pair(int x, int y) {
                this.x = x;
                this.y = y;
                this.z = 0;
            }

            public Pair(int x, int y, int z) {
                this.x = x;
                this.y = y;
                this.z = z;
            }

            public boolean equals(Object o) {
                if (this == o) {
                    return true;
                }

                if (o == null || getClass() != o.getClass()) {
                    return false;
                }
                BBeautifulNumbers.Pair pair = (BBeautifulNumbers.Pair) o;
                return x == pair.x && y == pair.y && z == pair.z;
            }

            public int hashCode() {
                int result = Integer.hashCode(x);
                result = 31 * result + Integer.hashCode(y);
                result = 31 * result + Integer.hashCode(z);
                return result;
            }

        }

    }

    static class inputClass {
        BufferedReader br;
        StringTokenizer st;

        public inputClass(InputStream in) {

            br = new BufferedReader(new InputStreamReader(in));
        }

        public String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

    }
}

