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 C (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include<math.h>
#include<stdlib.h>

int main(void){
	int N,i,P,max,min;
	//int P[200000];
	int count = 0;
	scanf("%d", &N);
	/*
	min = 99;
	for(i = 0; i < N; i++) {
		scanf("%d", &P[i]);
		if(P)
	}
	max = min = P[N - 1];
	for (i = N - 1; i > 0; i--) {
	}
	*/
	
	scanf("%d", &P);
	min = P;
	for (i = 1; i < N; i++) {
		scanf("%d", &P);
		if (min >= P) {
			count++;
			min = P;
		}
	}
	printf("%d \n", count+1);

	return 0;
}