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: Accepted)

#include <stdio.h>

int main(){
    int N;
    scanf("%d", &N);
    
    int i,j;
    int P[200000+1] = {0};
    // sum[i] : 0~i 番目の中での最小値
    int sum[200000+1] = {0};
    sum[0]=200000+1;
    for( i = 1; i <= N; i++ ){
        scanf("%d", &P[i]);

        if( P[i] < sum[i-1] ) sum[i]=P[i];
        else sum[i]=sum[i-1];
    }

    int flag=0;
    for( i = 1; i <= N; i++ ){
        if( P[i] > sum[i-1] ) flag++;
    }

    printf("%d\n", N-flag);

    return 0;
}