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.*;

public class B {
    Random random = new Random(751454315315L + System.currentTimeMillis());

    public void solve() throws IOException {
        int q = nextInt();
        while (q-- > 0) {
            int n = nextInt();
            int[] p = new int[n + 2];
            p[n + 1] = p[0] = 999999999;
            for (int i = 0; i < n; i++) {
                p[i + 1] = nextInt() - 1;
            }
            boolean[] used = new boolean[n + 1];
            used[0] = true;
            int lastUsed = 0;
            int i = 0;
            while (p[i] != 0) {
                i++;
            }
            int j = i;
            i--;
            j++;
            ArrayList<Integer> ans = new ArrayList<>();
            ans.add(0);
            int len = 0;
            int maxused = 0;
            while (i > 0 || j < n + 1) {
                int next;
                if (p[i] < p[j]) {
                    next = p[i];
                    i--;
                } else {
                    next = p[j];
                    j++;
                }
                used[next] = true;
                len++;
                maxused = Math.max(maxused, next);
                while (used[lastUsed + 1]) {
                    lastUsed++;
                }
                if (maxused == lastUsed) {
                    ans.add(len);
                }
            }
            used = new boolean[n];
            for (int x : ans) {
                used[x] = true;
            }
            for (int x = 0; x < n; x++) {
                out.print(used[x] ? "1" : "0");
            }
            out.println();
        }
    }

    public void run() {
        try {
            br = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    private void shuffle(int[] s) {
        for (int i = 0; i < s.length; ++i) {
            int j = random.nextInt(i + 1);
            int t = s[i];
            s[i] = s[j];
            s[j] = t;
        }
    }

    class Pair implements Comparable<Pair> {
        int f;
        int s;

        public Pair(int f, int s) {
            this.f = f;
            this.s = s;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Pair pair = (Pair) o;
            return f == pair.f &&
                    s == pair.s;
        }

        @Override
        public int hashCode() {
            return Objects.hash(f, s);
        }

        @Override
        public int compareTo(Pair o) {
            if (f == o.f) {
                return s - o.s;
            }
            return f - o.f;
        }
    }

    BufferedReader br;
    StringTokenizer in;
    PrintWriter out;

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

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

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

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

    public int[] nextArr(int n) throws IOException {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextInt();
        }
        return res;
    }

    public static void main(String[] args) throws IOException {
        Locale.setDefault(Locale.US);
        new B().run();
    }
}