Problem:
Score : 300 points

Problem Statement

Given is a permutation P_1, \ldots, P_N of 1, \ldots, N.
Find the number of integers i (1 \leq i \leq N) that satisfy the following condition:

For any integer j (1 \leq j \leq i), P_i \leq P_j.

Constraints

1 \leq N \leq 2 \times 10^5

P_1, \ldots, P_N is a permutation of 1, \ldots, N.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
P_1 ... P_N

Output

Print the number of integers i that satisfy the condition.

Sample Input 1

5
4 2 5 1 3

Sample Output 1

3

i=1, 2, and 4 satisfy the condition, but i=3 does not - for example, P_i > P_j holds for j = 1.

Similarly, i=5 does not satisfy the condition, either. Thus, there are three integers that satisfy the condition.

Sample Input 2

4
4 3 2 1

Sample Output 2

4

All integers i (1 \leq i \leq N) satisfy the condition.

Sample Input 3

6
1 2 3 4 5 6

Sample Output 3

1

Only i=1 satisfies the condition.

Sample Input 4

8
5 7 4 2 6 8 1 3

Sample Output 4

4

Sample Input 5

1
1

Sample Output 5

1

Solution in Rust (1.15.1) (judge verdict: Accepted)

macro_rules! get_one {
    ($t:ty) => {{
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().parse::<$t>().unwrap()
    }};

    ($t:ty; $n:expr) => {
        (0..$n).map(|_| get_one!($t)).collect::<Vec<_>>()
    };
}

macro_rules! get_vec {
    ($t:ty) => {
        {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            line.split_whitespace()
                .map(|t| t.parse::<$t>().unwrap())
                .collect::<Vec<_>>()
        }
    };
    ($t:ty; $n:expr) => {
        (0..$n).map(|_| get_vec!($t)).collect::<Vec<_>>()
    };
}

fn main() {
    let n = get_one!(usize);
    let ps = get_vec!(usize);


    let mut ps_idx: Vec<_> = ps.into_iter().enumerate().collect();
    ps_idx.sort_by_key(|x| x.1);

    let mut count = 0;
    let mut max_idx = n;
    for &(idx, _) in ps_idx.iter() {
        if max_idx > idx {
            count += 1;
            max_idx = idx;
        }
    }

    println!("{}", count)

    // for i in ps_idx.iter() {
    //     match ps.iter().position(|&x| x == i) {
    //         Some(pos) => {
    //             count += 1;
    //             ps = ps.into_iter().take(pos).collect();
    //         },
    //         None => continue ,
    //     }
    // }
    //
    // println!("{}", count)

}