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.IOException;
import java.io.InputStream;
import java.util.*;
public class sol {
    public static void main(String args[])throws Exception {
        FastReader in = new FastReader(System.in);    
        int t = in.nextInt(),i,j;
        StringBuilder sb = new StringBuilder();
        while(t-->0) { 
            int n=in.nextInt(),x,max=-1,min=Integer.MAX_VALUE;
            HashMap<Integer,Integer> map=new HashMap<>();
            for(i=1;i<=n;i++) {
                x=in.nextInt();
                map.put(x,i);      
            }
            for(i=1;i<=n;i++) {
                x=map.get(i);
                max=Math.max(max,x);
                min=Math.min(min,x);
                if((max-min+1)==i)
                    sb.append("1");
                else
                    sb.append("0");
            }
            sb.append("\n");
        }
        System.out.print(sb);
    }
}
class FastReader {
    byte[] buf = new byte[2048];
    int index, total;
    InputStream in;
    FastReader(InputStream is) {
        in = is;
    }
    int scan() throws IOException {
        if (index >= total) {
            index = 0;
            total = in.read(buf);
            if (total <= 0) {
                return -1;
            }
        }
        return buf[index++];
    }
    String next() throws IOException {
        int c;
        for (c = scan(); c <= 32; c = scan()) ;
        StringBuilder sb = new StringBuilder();
        for (; c > 32; c = scan()) {
            sb.append((char) c);
        }
        return sb.toString();
    }
    String nextLine() throws IOException {
        int c;
        for (c = scan(); c <= 32; c = scan()) ;
        StringBuilder sb = new StringBuilder();
        for (; c != 10 && c != 13; c = scan()) {
            sb.append((char) c);
        }
        return sb.toString();
    }
    char nextChar() throws IOException {
        int c;
        for (c = scan(); c <= 32; c = scan()) ;
        return (char) c;
    }
    int nextInt() throws IOException {
        int c, val = 0;
        for (c = scan(); c <= 32; c = scan()) ;
        boolean neg = c == '-';
        if (c == '-' || c == '+') {
            c = scan();
        }
        for (; c >= '0' && c <= '9'; c = scan()) {
            val = (val << 3) + (val << 1) + (c & 15);
        }
        return neg ? -val : val;
    }
    long nextLong() throws IOException {
        int c;
        long val = 0;
        for (c = scan(); c <= 32; c = scan()) ;
        boolean neg = c == '-';
        if (c == '-' || c == '+') {
            c = scan();
        }
        for (; c >= '0' && c <= '9'; c = scan()) {
            val = (val << 3) + (val << 1) + (c & 15);
        }
        return neg ? -val : val;
    }
}

/*
50 -> 433
100 -> 1183
5000 -> 388797
10000 -> 1092257
*/
