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.swing.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.security.cert.CollectionCertStoreParameters;
import java.util.*;



import static javax.swing.UIManager.get;
import static javax.swing.UIManager.getString;


public class Main {


    static class Pair implements Comparable<Pair>{
        int x=0;
        int y=0;
        public Pair(int x1,int y1)
        { x=x1;
            y=y1;
        }


        @Override
        public int compareTo(Pair o) {
            return this.x-o.x;
        }
    }

    static boolean checkPallindrome(int n)
    {    ArrayList<Integer> list = new ArrayList<>();
        while(n>0)
        { list.add(n%10);
            n/=10;
        }
        int low=0,high=list.size()-1;
        while(low<=high)
        {     if(list.get(low)!=list.get(high))
            return false;
            low++;
            high--;
        }
        return true;
    }


    static boolean check(String s)
    {

        int low=0,high=s.length()-1;
        while(low<=high)
        {     if(s.charAt(low)!=s.charAt(high))
            return false;
            low++;
            high--;
        }
        return true;
    }

    static  int min=Integer.MAX_VALUE;
    public static void main(String[] args) throws IOException {
        FastReader sc = new FastReader();
//(1)very very important**(never take the first problem for granted, always check the test cases) take 5 minutes more and check the edge cases
        // 5 minutes will not decreases rating as much as a wrong submission does it is easy u just think with an open mind and u will surely get the answer
        //(2)let ur brain consume the problem don't just jump to the solution. after reading the problem take a pause 1 minute
        //(3)go through the example test cases and also at least two of ur own test cases.Think of testcases which are difficult(edge cases).dry run ur concept

        int t = sc.nextInt();
        while (t-- > 0) {
   int n =sc.nextInt();
    int[] a = sc.readArray(n);
int[] b = new int[n+1];
for(int i=0;i<n;i++)
    b[a[i]]=i;

long sum=0L;
long min =10000007L;
StringBuilder sb = new StringBuilder();
for(int i=1;i<=n;i++)
{  sum+=b[i];
    min=Math.min(min,b[i]);
    if((sum-i*min)==(1L*i*(i-1))/2)
        sb.append('1');
    else sb.append('0');
}

System.out.println(sb);

        }
    }








//Things to check when u r getting wrong answer
// syntax of conditional operator y=(x==1)?1:0;
// 1-  check the flow of the code
//2- If ur stuck read the problem once again
//3- before submitting always check the output format of ur code
//4- don't check standings until problem B is done
//5- if u r thinking ur concept is correct but still u r getting wrong answer try to implement it in another way
//6- By default, java interpret all numeral literals as 32-bit integer values.
// If you want to explicitely specify that this is something bigger then 32-bit integer you should use suffix L for long values.    example   long a = 600851475143L

//All the functions



    static void debug(String s)
    { System.out.println(s);

    }

    //collections.sort use merge sort instead of quick sort but arrays.sort use quicksort whose worst time complexity is O(n^2)
    static int[] sort(int[] a)
    {         ArrayList<Integer> list = new ArrayList<>();
        for(int i=0;i<a.length;i++)
            list.add(a[i]);

        Collections.sort(list);
        int ind=0;
        for(int x:list)
            a[ind++]=x;
        return a;

    }

    //function to print an array for debugging
    static void print(int[] a) {
        for (int i = 0; i < a.length; i++)
            System.out.print(a[i] + " ");
        System.out.println();
    }

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


    //normal gcd function, always put the greater number as a and the smaller number as b
    static int gcd(int a, int b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);

    }

    static int lcm(int a,int b)
    {
        return (a*b)/gcd(a,b);
    }


    //to find gcd and lcm for numbers of long data type
    static long gcdl(long a, long b) {
        if (b == 0)
            return a;
        return gcdl(b, a % b);

    }

    static long lcml(long a,long b)
    {
        return (a*b)/gcdl(a,b);
    }


    //Input Reader to read faster input
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            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;
        }


        int[] readArray(int n)
        {  int[] a = new int[n];
            for(int i=0;i<n;i++)
                a[i]=nextInt();

            return a;
        }
    }



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