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 (Clang 3.8.0) (judge verdict: Accepted)

#include <stdio.h>

#define NUMBER 200000

int main(void)
{
    int n;
    int p[NUMBER];
    int i;
    int c = 1;
    
    scanf("%d",&n);
    for (i = 0;i < n;i++){
        scanf("%d",&p[i]);
        }
    
    int f[n];
    
    f[0] = p[0];
    for (i = 1;i < n;i++){
        if (p[i] <= f[i-1]){
            f[i] = p[i];
            c++;
        }
        else {
            f[i] = f[i-1];
        }
    }
    
    printf("%d\n",c);
    return 0;
}