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.util.*;
import java.io.*;
import static java.lang.Math.*;

public class Main implements Runnable
{
    boolean multiple = true;

    void solve() throws Exception
    {
        int n = sc.nextInt();
        int[] p = new int[n];
        for (int I = 0; I < n; I++)
            p[I] = sc.nextInt() ;
        boolean[] left = new boolean[n + 1];
        boolean seenOne = false;
        int L = -1, R = -1;
        for (int i = 0; i < n; i++)
        {
            if (p[i] == 1)
            {
                seenOne = true;
                L = R = i;
            }
            left[p[i]] = !seenOne;
        }

        boolean[] seen = new boolean[n + 1];
        seen[1] = true;

        for (int m = 1; m <= n; m++)
        {
//            System.out.println();
//            System.out.println();
//            System.out.println(m);
            while (!seen[m])
            {
                if (left[m])
                {
                    L--;
                    seen[p[L]] = true;
                }
                else
                {
                    R++;
                    seen[p[R]] = true;
                }
            }
//            System.out.println(L + ", " + R);
            System.out.print((R - L + 1 == m) ? 1 : 0);
        }
        System.out.println();
    }

    long inv(long a, long b)
    {
        return 1 < a ? b - inv(b % a, a) * b / a : 1;
    }

    long gcd(long a, long b)
    {
        return a == 0 ? b : gcd(b % a, a);
    }

    class Node implements Comparable<Node>
    {
        int idx, priority;

        Node(int i, int p)
        {
            idx = i;
            priority = p;
        }

        @Override
        public int compareTo(Node o)
        {
            return Integer.compare(priority, o.priority);
        }
    }

    void print(int[] arr)
    {
        for (int i = 0; i < arr.length; i++)
            System.out.print(arr[i] + " ");
        System.out.println();
    }

    @Override
    public void run()
    {
        try
        {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
            sc = new FastScanner(in);
            if (multiple)
            {
                int q = sc.nextInt();
                for (int i = 0; i < q; i++)
                    solve();
            }
            else
                solve();
        }
        catch (Throwable uncaught)
        {
            Main.uncaught = uncaught;
        }
        finally
        {
            out.close();
        }
    }

    public static void main(String[] args) throws Throwable
    {
        Thread thread = new Thread(null, new Main(), "", (1 << 26));
        thread.start();
        thread.join();
        if (Main.uncaught != null) {
            throw Main.uncaught;
        }
    }

    static Throwable uncaught; BufferedReader in; FastScanner sc; PrintWriter out;
}

class FastScanner
{
    BufferedReader in;
    StringTokenizer st;

    public FastScanner(BufferedReader in)
    {
        this.in = in;
    }

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

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

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

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