Problem:
Score : 200 points

Problem Statement

We have a permutation p = {p_1,\ p_2,\ ...,\ p_n} of {1,\ 2,\ ...,\ n}.

Print the number of elements p_i (1 < i < n) that satisfy the following condition:

p_i is the second smallest number among the three numbers p_{i - 1}, p_i, and p_{i + 1}.

Constraints

All values in input are integers.

3 \leq n \leq 20

p is a permutation of {1,\ 2,\ ...,\ n}.

Input

Input is given from Standard Input in the following format:

n
p_1 p_2 ... p_n

Output

Print the number of elements p_i (1 < i < n) that satisfy the condition.

Sample Input 1

5
1 3 5 4 2

Sample Output 1

2

p_2 = 3 is the second smallest number among p_1 = 1, p_2 = 3, and p_3 = 5. Also, p_4 = 4 is the second smallest number among p_3 = 5, p_4 = 4, and p_5 = 2. These two elements satisfy the condition.

Sample Input 2

9
9 6 3 2 5 8 7 4 1

Sample Output 2

5

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
 
int main() {
  int n;
  int p[20];
  int a = 0;
  scanf("%d", &n);
  for(int i = 0; i < n; i++) {
    scanf("%d", &p[i]);
  }
  for(int i = 0; i < n - 2; i++) {
    for(int j = i + 1; i < n - 1; j++) {
      if(p[i] > p[j]) {
        break;
      }
      for(int k = j + 1; k < n; k++) {
        if(p[j] > p[k]) {
          break;
        } 
        a = a + 1;
      }
    }
  }
  for(int i = n; i >= 2; i--) {
    for(int j = i - 1; i >= 1; j--) {
      if(p[i] > p[j]) {
        break;
      }
      for(int k = j - 1; k >= 0; k--) {
        if(p[j] > p[k]) {
          break;
        } 
        a = a + 1;
      }
    }
  }
  printf("%d", a);
  return 0;
}