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.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Set;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

public class A {
    static ArrayList<Integer> [] adj;
    static ArrayList<Integer> temp;
    static boolean[] vis = new boolean[(int)1e5];

    static boolean check(int[] a,int l,int r,int v){
        while(l<r){
            int mid=(l+r)/2;
            if(a[mid]>v)return false;
            else if(a[mid]<v)l++;
            else r--;
        }
        return true;
    }

    public static void main(String[] args) throws IOException {
        FastScanner s=new FastScanner();
        int mod = (int) 1e9+7;
      int t = s.nextInt();
//        int t = 1;
        while(t-->0) {
            int n = s.nextInt();
            int[] index = new int[n+1];
            int[] b = new int[n+1];
            int l=0;
            for(int i=1;i<=n;i++){
                int p = s.nextInt();
                index[p] =i;
                b[i]=p;
                if(p==1)l=i;
            }
            StringBuilder se = new StringBuilder("1");
            int max=index[1],min=index[1];
//            System.out.println(max+" "+min);
            for(int i=2;i<=n;i++){
                max=Math.max(index[i],max);
                min=Math.min(index[i],min);
//                System.out.println(max+" "+min);
                if(Math.abs(max-min)<i)se.append(1);
                else se.append(0);
            }
            System.out.println(se);
        }
    }
    static void dfs(int i){
        vis[i]=true;
        temp.add(i);
        List<Integer> Y = adj[i];
        for(int x:Y){
            if(!vis[x])dfs(x);
        }
    }

    static int lower_bound(int[] arr, int x) {
        int low_limit = 0, high_limit = arr.length, mid = -1;
        while (low_limit < high_limit) {
            mid = (low_limit + high_limit) / 2;
            if (arr[mid] >= x) {
                high_limit = mid;
            } else {
                low_limit = mid + 1;
            }
        }
        return low_limit + 1;
    }

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

    static int gcd(int a, int b) throws IOException {
        return (b == 0) ? a : gcd(b, a % b);
    }

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

    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 void sort(long[] a) {
        ArrayList<Long> l = new ArrayList<>();
        for (long i : a)
            l.add(i);
        Collections.sort(l);
        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 nextLong() {
            return Long.parseLong(next());
        }
    }
}
