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>
#define NUM 200000

int main(){
  int n;
  int p[NUM];
  int i,j,ans = 1;

  /*入力*/
  scanf("%d",&n);
  for(i = 0;i < n;i++){
    scanf("%d",&p[i]);
  }

  /*調べる*/
  int s = 0;
  for(i = 1;i < n;i++){
    s = 0;
    for(j = 0;j < i;j++){
      //printf("比較: %d %d\n",p[i],p[j]);
      if(p[i] >= p[j]){
	break;
      }
      s++;
    }
    if(s == i){
      //printf("%d\n",p[i]);
      ans++;
    }
  }

  printf("%d\n",ans);

  return 0;
}