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. 
//package cf604;

import java.io.*;
import java.util.*;

public class beautifulNum {
    static BufferedReader in;

    public static void solve() throws IOException {
        int t = nextInt();
        while(t-- > 0){
            int n = nextInt();
            Integer[] arr = nextArray(n);
            Pair[] pair = new Pair[n];

            for (int i = 0; i < n; i++) {
                pair[i] = new Pair(arr[i], i+1);
            }

            Arrays.sort(pair);

            StringBuilder ans = new StringBuilder();

            int pmin = pair[0].pos;
            int pmax = pair[0].pos;

            for(Pair p : pair){
                int m = p.num;
                int pm = p.pos;
                pmin = Math.min(pmin,pm );
                pmax = Math.max(pmax,pm);

                if(pmax - pmin + 1 == m)
                    ans.append("1");
                else
                    ans.append("0");

            }

            System.out.println(ans.toString());

        }

    }

    public static void main(String[] args) throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        solve();
    }

    static int nextInt() throws IOException {
        return Integer.parseInt(in.readLine().trim());
    }

    static long nextLong() throws IOException {
        return Long.parseLong(in.readLine().trim());
    }

    static Integer[] nextArray(int n) throws IOException {
        StringTokenizer st = new StringTokenizer(in.readLine());
        Integer[] tmp = new Integer[n];
        for (int i = 0; i < tmp.length; i++) {
            tmp[i] = Integer.parseInt(st.nextToken());
        }

        return tmp;
    }

}

class Pair implements Comparable<Pair>{
    int num;
    int pos;

    public Pair(int num, int pos) {
        this.num = num;
        this.pos = pos;
    }

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

