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(void){

    int N = 0;

    scanf("%d", &N);

    int count = N;
    int min = 9999999;

    int P[N];
    int check[N];

    for (int i = 0; i < N; i++){

        int flag = 0;
        check[i] = 0;

        scanf("%d", &P[i]);

        for (int j = i; j >= 0; j--){

            if(min > P[i]){

                min = P[i];
                break;
            }

            if (P[j] == P[i] + 1 && check == 1){

                count--;
                break;
            }

            if (P[i] > P[j]){

                count--;
                check[i] = 1;
                break;
            }
        } 
    }

    printf("%d", count);

    return 0;
}