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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		//入力（Nと配列）
		Scanner sc = new Scanner(System.in);
		int N = Integer.parseInt(sc.nextLine());
		int[] PList = new int[N];

		//左に自分より小さい数があればNG
			//今まで出てきた最も小さい数を記憶する。比較する。
		//配列の最初は比較対象がない。
		PList[0] = sc.nextInt();
		int minNumber = PList[0];
		int cnt=1;

		for(int i=1;i<N;i++) {
			PList[i] = sc.nextInt();
			if(minNumber>=PList[i]) {
				minNumber = PList[i];
				cnt++;
			}else {
			}
		}
		System.out.println(cnt);
	}
}