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 javax.print.DocFlavor;
import java.io.*;
import java.math.BigInteger;
import java.util.*;
import java.util.concurrent.ThreadLocalRandom;

import static java.lang.Math.*;

@SuppressWarnings("Duplicates")

public class solveKEK {
    FastScanner in;
    PrintWriter out;
    boolean systemIO = true, multitests = true;
    long INF = Integer.MAX_VALUE / 2, mod = (long) (1e9 + 7);


    void solve() {
        int n = in.nextInt();
        int id[] = new int[n];
        for (int i = 0; i < n; i++) {
            id[in.nextInt() - 1] = i;
        }
        int l = n, r = 0;
        for (int i = 0; i < n; ++i) {
            l = min(l, id[i]);
            r = max(r, id[i]);
            if (r - l == i) {
                out.print('1');
            } else {
                out.print('0');
            }
        }
        out.println();
    }

    class hash {
        long mod = (int) (1e9 + 7), p = 137;
        int n;
        long[] pref, powp;

        long make(long a, long b) {
            return (a % b + b) % b;
        }

        hash(String a) {
            n = a.length();
            pref = new long[n + 1];
            powp = new long[n + 1];
            powp[0] = 1;
            for (int i = 0; i < n; ++i) {
                pref[i + 1] = make(pref[i] * p + a.charAt(i), mod);
                powp[i + 1] = make(powp[i] * p, mod);
            }
        }

        long get(int l, int r) {
            if (l > r)
                return 0;
            return make(pref[r + 1] - make(pref[l] * powp[r - l + 1], mod), mod);
        }


    }

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

    void shuffleArray(long[] ar) {
        Random rnd = ThreadLocalRandom.current();
        for (int i = ar.length - 1; i > 0; i--) {
            int index = rnd.nextInt(i + 1);
            long a = ar[index];
            ar[index] = ar[i];
            ar[i] = a;
        }
    }

    long[] readArray(int size_) {
        long[] ar = new long[size_];
        for (int i = 0; i < size_; i++) {
            ar[i] = in.nextLong();
        }
        return ar;
    }

    void printArray(long[] ar) {
        for (long k : ar) {
            System.out.print(k + " ");
        }
        System.out.println();
    }

    void reverseArray(long[] ar) {
        for (int i = 0, j = ar.length - 1; i < j; i++, j--) {
            long a = ar[i];
            ar[i] = ar[j];
            ar[j] = a;
        }
    }

    String reverseString(String x) {
        StringBuilder l = new StringBuilder(x);
        l.reverse();
        return l.toString();
    }

    class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        FastScanner(File f) {
            try {
                br = new BufferedReader(new FileReader(f));
            } catch (FileNotFoundException e) {
                e.printStackTrace();
            }
        }

        FastScanner(InputStream f) {
            br = new BufferedReader(new InputStreamReader(f));
        }

        String nextLine() {
            try {
                return br.readLine();
            } catch (IOException e) {
                return null;
            }
        }

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

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

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

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

    }

    //String.format("%8.3f", ans1).replace(',', '.')
    private void run() throws IOException {
        if (systemIO) {
            in = new solveKEK.FastScanner(System.in);
            out = new PrintWriter(System.out);
        } else {
            in = new solveKEK.FastScanner(new File("birch.in"));
            out = new PrintWriter(new File("birch.out"));
        }
        for (int t = multitests ? in.nextInt() : 1; t-- > 0; )
            solve();

        out.close();
    }

    public static void main(String[] arg) throws IOException {
        new solveKEK().run();
    }

}

