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: Time Limit Exceeded)

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

int main(){
	int n, i, j, min, max;
	int *num = NULL;
	int cnt = 0;
	
	scanf("%d", &n);
	
	num = (int *)malloc(sizeof(int)*n);
	
	for(i=0; i<n; ++i){
		
		scanf("%d", &num[i]);
		
		// min, maxの初期化
		if(i==0){
			min = num[0];
			max = num[0];
			++cnt;
			continue;
		}
		
		if(min >= num[i]){
			min = num[i];
			++cnt;
			continue;
		}
		
		if(max < num[i]){
			max = num[i];
			continue;
		}
		
		for(j=0; j<i; ++j){
			if(num[i]>num[j])
			break;
		}
		if(i==j){
			++cnt;
		}
	}
	printf("%d", cnt);
	
	free(num);
	
	return 0;
}