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. 
/**
 * BaZ :D
 */

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

public class Main
{
    static MyScanner scan;
    static PrintWriter pw;
    static long MOD = 1_000_000_007;
    static long INF = 1_000_000_000_000_000_000L;
    static long inf = 2_000_000_000;
    public static void main(String[] args) {
        new Thread(null, null, "BaZ", 1 << 27) {
            public void run() {
                try {
                    solve();
                } catch (Exception e) {
                    e.printStackTrace();
                    System.exit(1);
                }
            }
        }.start();
    }

    static void solve() throws IOException
    {
        //initIo(true);
        initIo(false);
        StringBuilder sb = new StringBuilder();
        int t = ni();
        while(t-->0) {
            int n = ni();
            int pos[] = new int[n+1];
            for(int i=0;i<n;++i) {
                int x = ni();
                pos[x] = i;
            }
            int max = Integer.MIN_VALUE , min = Integer.MAX_VALUE;
            for(int i=1;i<=n;++i) {
                max = max(max, pos[i]);
                min = min(min, pos[i]);
                int range_size = max - min + 1;
                if(range_size == i) {
                    sb.append("1");
                }
                else {
                    sb.append("0");
                }
            }
            sb.append("\n");
        }
        p(sb);
        pw.flush();
        pw.close();
    }
    static void initIo(boolean isFileIO) throws IOException {
        scan = new MyScanner(isFileIO);
        if(isFileIO) {
            pw = new PrintWriter("/Users/amandeep/Desktop/output.txt");
        }
        else {
            pw = new PrintWriter(System.out, true);
        }
    }
    static int ni() throws IOException
    {
        return scan.nextInt();
    }
    static long nl() throws IOException
    {
        return scan.nextLong();
    }
    static double nd() throws IOException
    {
        return scan.nextDouble();
    }
    static String ne() throws IOException
    {
        return scan.next();
    }
    static String nel() throws IOException
    {
        return scan.nextLine();
    }
    static void pl()
    {
        pw.println();
    }
    static void p(Object o)
    {
        pw.print(o+" ");
    }
    static void pl(Object o)
    {
        pw.println(o);
    }
    static void psb(StringBuilder sb)
    {
        pw.print(sb);
    }
    static void pa(String arrayName, Object arr[])
    {
        pl(arrayName+" : ");
        for(Object o : arr)
            p(o);
        pl();
    }
    static void pa(String arrayName, int arr[])
    {
        pl(arrayName+" : ");
        for(int o : arr)
            p(o);
        pl();
    }
    static void pa(String arrayName, long arr[])
    {
        pl(arrayName+" : ");
        for(long o : arr)
            p(o);
        pl();
    }
    static void pa(String arrayName, double arr[])
    {
        pl(arrayName+" : ");
        for(double o : arr)
            p(o);
        pl();
    }
    static void pa(String arrayName, char arr[])
    {
        pl(arrayName+" : ");
        for(char o : arr)
            p(o);
        pl();
    }
    static void pa(String listName, List list)
    {
        pl(listName+" : ");
        for(Object o : list)
            p(o);
        pl();
    }
    static void pa(String arrayName, Object[][] arr) {
        pl(arrayName+" : ");
        for(int i=0;i<arr.length;++i) {
            for(Object o : arr[i])
                p(o);
            pl();
        }
    }
    static void pa(String arrayName, int[][] arr) {
        pl(arrayName+" : ");
        for(int i=0;i<arr.length;++i) {
            for(int o : arr[i])
                p(o);
            pl();
        }
    }
    static void pa(String arrayName, long[][] arr) {
        pl(arrayName+" : ");
        for(int i=0;i<arr.length;++i) {
            for(long o : arr[i])
                p(o);
            pl();
        }
    }
    static void pa(String arrayName, char[][] arr) {
        pl(arrayName+" : ");
        for(int i=0;i<arr.length;++i) {
            for(char o : arr[i])
                p(o);
            pl();
        }
    }
    static void pa(String arrayName, double[][] arr) {
        pl(arrayName+" : ");
        for(int i=0;i<arr.length;++i) {
            for(double o : arr[i])
                p(o);
            pl();
        }
    }
    static class MyScanner
    {
        BufferedReader br;
        StringTokenizer st;
        MyScanner(boolean readingFromFile) throws IOException
        {
            if(readingFromFile) {
                br = new BufferedReader(new FileReader("/Users/amandeep/Desktop/input.txt"));
            }
            else {
                br = new BufferedReader(new InputStreamReader(System.in));
            }
        }
        String nextLine()throws IOException
        {
            return br.readLine();
        }
        String next() throws IOException
        {
            if(st==null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }
        int nextInt() throws IOException
        {
            return Integer.parseInt(next());
        }
        long nextLong() throws IOException
        {
            return Long.parseLong(next());
        }
        double nextDouble() throws IOException
        {
            return Double.parseDouble(next());
        }
    }
}