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>
#include<stdlib.h>
#include<string.h>
int main(){
  int i,j,n,p[500000],count=0,ans=0;
  scanf("%d",&n);
  for(i=0;i<n;i++){
    scanf("%d",&p[i]);
  }
  for(i=1;i<n;i++){
    count=0;
    for(j=0;j<i;j++){
      if(p[j]>=p[i]){
        count++;
        //printf("p[i]=%d,p[j]=%d,count=%d,ans=%d\n",p[i],p[j],count,ans);
        if(count==i){
          count=0;
          ans++;
        }
      }
    }
  }
  printf("%d\n",ans+1);
  return 0;
}