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 sun.nio.cs.KOI8_U;

import java.io.*;
import java.util.*;
import java.math.*;
import java.lang.*;

public class AD implements Runnable {


    public void run() {
        InputReader sc = new InputReader();
        PrintWriter out = new PrintWriter(System.out);
        int i=0,j=0,k=0;
        int t=0;
        t=sc.nextInt();
        for(int testcase = 0; testcase < t; testcase++)
        {
            int n=sc.nextInt();
            int arr[]=new int[n+1];
            for (i=0;i<n;i++)
            {
                arr[sc.nextInt()]=i;
            }
            int max=Integer.MIN_VALUE;
            int min=Integer.MAX_VALUE;

            for (i=1;i<=n;i++)
            {
                min=Math.min(arr[i],min);
                max=Math.max(arr[i],max);
                if (max-min+1==i)
                    out.print("1");
                else
                    out.print("0");
            }
            out.println();
        }

//================================================================================================================================
        out.flush();
        out.close();
    }


    //================================================================================================================================
    public static int[] sa(int n,InputReader sc)
    {
        int arr[]=new int[n];
        for (int i=0;i<n;i++)
            arr[i]=sc.nextInt();
        return arr;
    }
    public static long gcd(long a,long b){
        return (a%b==0l)?b:gcd(b,a%b);
    }
    private static long lcm(long a, long b)
    {
        return a * (b / gcd(a, b));
    }
    public int egcd(int a, int b) {
        if (a == 0)
            return b;

        while (b != 0) {
            if (a > b)
                a = a - b;
            else
                b = b - a;
        }

        return a;
    }



    public int countChar(String str, char c)
    {
        int count = 0;
        for(int i=0; i < str.length(); i++)
        {    if(str.charAt(i) == c)
            count++;
        }
        return count;
    }
    static int binSearch(Integer[] arr, int number){
        int left=0,right=arr.length-1,mid=(left+right)/2,ind=0;
        while(left<=right){
            if(arr[mid]<=number){
                ind=mid+1;
                left=mid+1;
            }
            else
                right=mid-1;
            mid=(left+right)/2;
        }
        return ind;
    }
    static int binSearch(int[] arr, int number){
        int left=0,right=arr.length-1,mid=(left+right)/2,ind=0;
        while(left<=right){
            if(arr[mid]<=number){
                ind=mid+1;
                left=mid+1;
            }
            else
                right=mid-1;
            mid=(left+right)/2;
        }
        return ind;
    }
    static class Pair
    {
        int a,b;
        Pair(int aa,int bb)
        {
            a=aa;
            b=bb;
        }
        String get()
        {
            return a+" "+b;
        }
        String getrev()
        {
            return b+" "+a;
        }
    }

    static boolean isPrime(long n) {
        if(n < 2) return false;
        if(n == 2 || n == 3) return true;
        if(n%2 == 0 || n%3 == 0) return false;
        long sqrtN = (long)Math.sqrt(n)+1;
        for(long i = 6L; i <= sqrtN; i += 6) {
            if(n%(i-1) == 0 || n%(i+1) == 0) return false;
        }
        return true;
    }
    static long factorial(long n)
    {
        if (n == 0)
            return 1;
        return n*factorial(n-1);
    }
//================================================================================================================================

    static class InputReader
    {
        BufferedReader br;
        StringTokenizer st;

        public InputReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                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 nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }

    public static void main(String args[]) throws Exception {
        new Thread(null, new AD(),"Main",1<<27).start();
    }
}
