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>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include <time.h>
typedef long long llint;
typedef long double ldouble;
// 小大
int compare(const void *a, const void *b){return *(llint *)a - *(llint *)b;}
typedef struct{
    llint aa;
    llint bb;
}frequent;
// 小大
int cmp( const void *p, const void *q ) {
    return ((frequent*)p)->bb - ((frequent*)q)->bb;
}
/*2つのうち小さい数を返す*/
llint min2(llint a,llint b){return a>=b?b:a;}

int main(void){
    llint n,i,ans=0,most;
    scanf("%lld",&n);
    llint p[n];
    for(i=0;i<n;i++){
        scanf("%lld",&p[i]);
        if(i==0){
            most=p[0];
            ans++;
            continue;
        }
        most=min2(most,p[i]);
        if(most==p[i]) ans++;
    }



    printf("%lld\n",ans);
    return 0;
}