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>

int main(int argc, char const *argv[])
{
    int p_size = 0;
    int i = 0;
    int j = 0;
    int count = 0;
    int fill_count = 0;

    scanf("%d", &p_size);

    int p[p_size];

    for (i = 0; i < p_size; i++)
    {
        p[i] = 0;
        scanf("%d", &p[i]);
    }

    for (i = 0; i < p_size; i++)
    {
        for (j = 0; j <= i; j++)
        {
            if (p[i] <= p[j])
            {
                count++;
            }
            else
            {
                break;
            }
        }
        if (i + 1 == count)
        {
            fill_count++;
        }
        count = 0;
    }

    printf("%d\n", fill_count);

    return 0;
}