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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
//_________________________________________________________________
public class Main {

    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
//        Scanner sc= new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int t=sc.nextInt();
        while (t-->=1) {
            int n = sc.nextInt();
            int a[] = sc.readArray(n);
            int pos[]= new int[n];
            for (int i=0;i<n;i++){
                pos[a[i]-1]=i;
            }
            int min=n,max=0;
            StringBuilder ans=new StringBuilder();
            for (int i=0;i<n;i++){
                min=Math.min(min,pos[i]);
                max=Math.max(max,pos[i]);
                if (max-min==i){
                    ans.append('1');
                }
                else ans.append('0');
            }
            System.out.println(ans.toString());


        }

               out.flush();
    }

    //------------------------------------if-------------------------------------------------------------------------------------------------------------------------------------------------

    static void sort(int[] a) {
        ArrayList<Integer> l = new ArrayList<>();
        for (int i : a) l.add(i);
        Collections.sort(l);
        for (int i = 0; i < a.length; i++) a[i] = l.get(i);
    }

    static String sortString(String s) {
        char temp[] = s.toCharArray();
        Arrays.sort(temp);
        return new String(temp);
    }

    static class Pair implements Comparable<Pair> {
        int a, b;

        public Pair(int a, int b) {
            this.a = a;
            this.b = b;
        }

        public int compareTo(Pair other) {
            if (this.a == other.a) return other.b > this.b ? -1 : 1;
            else if (this.a > other.a) return 1;
            else return -1;
        }

    }
    static int[] frequency(String s){
        int fre[]= new int[26];
        for (int i=0;i<s.length();i++){
            fre[s.charAt(i)-'a']++;
        }
        return fre;
    }

    static int mod = (int) 1e9 + 7;

    static long mod(long x) {
        return ((x % mod + mod) % mod);
    }

    static long add(long x, long y) {
        return mod(mod(x) + mod(y));
    }

    static long mul(long x, long y) {
        return mod(mod(x) * mod(y));
    }


    static int[] find(int n, int start, int diff) {
        int a[] = new int[n];
        a[0] = start;
        for (int i = 1; i < n; i++) a[i] = a[i - 1] + diff;
        return a;
    }

    static void swap(int a, int b) {
        int c = a;
        a = b;
        b = c;
    }

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

    static boolean sorted(int a[]) {
        int n = a.length;
        boolean flag = true;
        for (int i = 0; i < n - 1; i++) {
            if (a[i] > a[i + 1]) flag = false;
        }
        if (flag) return true;
        else return false;
    }

    public static int findlog(long n) {
        if (n == 0)
            return 0;

        if (n == 1)
            return 0;
        if (n == 2)
            return 1;
        double num = Math.log(n);
        double den = Math.log(2);
        if (den == 0)
            return 0;
        return (int) (num / den);
    }

    public static long gcd(long a, long b) {
        if (b % a == 0)
            return a;
        return gcd(b % a, a);
    }

    public static int gcdInt(int a, int b) {

        if (b % a == 0)
            return a;

        return gcdInt(b % a, a);

    }

    static void sortReverse(int[] a) {
        ArrayList<Integer> l = new ArrayList<>();
        for (int i : a) l.add(i);
//        Collections.sort.(l);
        Collections.sort(l, Collections.reverseOrder());
        for (int i = 0; i < a.length; i++) a[i] = l.get(i);
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

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

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

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

        long[] readArrayLong(long n) {
            long[] a = new long[(int) n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

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

}