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: Time Limit Exceeded)

#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::*;

use std::io::*;
use std::str::FromStr;

fn read<T: FromStr>() -> T {
  let stdin = stdin();
  let stdin = stdin.lock();
  let token: String = stdin
    .bytes()
    .map(|c| c.expect("failed reading char") as char)
    .skip_while(|c| c.is_whitespace())
    .take_while(|c| !c.is_whitespace())
    .collect();
  token.parse().ok().expect("failed parsing")
}

fn main() {
  let n: usize = read::<usize>();

  let mut p: Vec<usize> = vec![];

  for _ in 0..n {
    p.push(read::<usize>());
  }

  let mut count: usize = 1;
  for i in 1..n {
    let pj = &p[..(i + 1)];
    if p[i] <= *(pj.iter().min().unwrap()) {
      count = count + 1;
    }
  }

  println!("{}", count);
}