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<string.h>
#include<math.h>

#define MAX(a, b) ((a) > (b) ? (a) : (b))
#define MIN(a, b) ((a) < (b) ? (a) : (b))

int main(void){
	int n;
	int i, j;
	int cnt = 0;
	scanf("%d",&n);
	int p[n];
	int flag[n];
	for(i=0;i<n;i++){
		scanf("%d",&p[i]);
		flag[i] = 0;
	}
	for(i=1;i<n;i++){
		for(j=i-1;j>=0;j--){
			if(p[i] < p[i-1] && flag[i-1]==0){
				break;
			}
			if(p[i] > p[j]){
				cnt++;
				flag[i] = 1;
//				printf(" %d %d\n",i ,j);
				break;
			}
			if(flag[j] == 0 && p[j] > p[i]){
				break;
			}
//			printf("%d %d\n",i ,j);
		}
	}
	printf("%d\n",n - cnt);
    return 0; 
}