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: Wrong Answer)

#include<stdio.h>
int main(){
    int a,b,c,d,i,j,k,count=0,counts=0;
    scanf("%d",&a);
    int con[a];
    scanf("%d",&con[a]);
//    for(i =0;i<a;i++){
//        scanf("%d",&con[i]);   
//    }
    for(j=0;j<a;j++){
        if(j==0){
            count++;
        }
        else{
            for(k=j;k>0;k--){
                if(con[j]<con[k-1]){
                    counts++;
                }
                else{
                    break;
                }
            }
            if(counts==j){
                count++;

            }
            counts=0;
        }
    }
    printf("%d",count);
}