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 second(int a, int b, int c)
{
  int i,max=a,min=a,checkmax[3],checkmin[3],kqua=0;
  checkmax[0]=1; checkmin[0]=1;
  for (i=1; i<3; i++)
    {
      checkmax[i]=0; checkmin[i]=0;
    }
  if (max<b) 
    {
      max=b; checkmax[0]=0; checkmax[1]=1; checkmax[2]=0;
    }
  if (min>b)
    {
      min=b; checkmin[0]=0; checkmin[1]=1; checkmin[2]=0;
    }
  if (max<c) 
    {
      max=c; checkmax[0]=0; checkmax[1]=0; checkmax[2]=1;
    }
  if (min>c) 
    {
      min=c; checkmin[0]=0; checkmin[1]=0; checkmin[2]=1;
    }
  for (i=0; i<3; i++)
    {
      if (checkmax[i]==0 && checkmin[i]==0) kqua=i;
    }
  if (kqua==0) return a;
  if (kqua==1) return b;
  if (kqua==2) return c;
}
int main()
{
  int p[20],n,i,check[100],kqua=0;
  scanf("%d", &n);
  for (i=0; i<n; i++) 
    {
      scanf("%d", &p[i]);
      check[i+1]=0;
    }
  for (i=1; i<n-1; i++)
   {
     //printf("%d\n", second(p[i-1],p[i],p[i+1]));
     check[second(p[i-1],p[i],p[i+1])]++;
   }
  for (i=0; i<n; i++)
    {
      if (check[i+1]!=0) kqua++;
    }
  printf("%d", kqua);
  return 0;
}