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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

/*
 * 152 C 
 */

#include <stdio.h>
#include <stdlib.h>

int
main(int argc, char **argv)
{
  int d_flag = 0;
  int n;
  int *a;
  int cnt = 0;
  int i, j;
  int min;

  if (argc == 2)
    d_flag = 1;

  scanf("%d", &n);
 
  a = (int *) malloc(n * sizeof(int));
  for (i = 0; i < n; i++)
    scanf("%d", a + i);

  min = a[0];

  if (d_flag) {
    for (i = 0; i < n; i++)
      printf("a[%d] = %d\n", i, a[i]);
  }

  for (i = 0; i < n; i++) {
    
    for (j = 0; j < i; j++) {
      if (a[j] < min)
	break;
    }
    
     if (i == j && a[j] <= min) {
       if (d_flag)
	 printf("i = %d, min = %d, new min = %d\n", i, min, a[i]);
       min = a[i];
       cnt++;
    } 
  }

  printf("%d\n", cnt);
}