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 Java7 (OpenJDK 1.7.0) (judge verdict: Accepted)

import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
    
    Scanner sc = new Scanner(System.in);
 
      //入力
      int N =sc.nextInt();//数字の個数
      int p[] =new int[N];//順列
      int count =0;
      
      //順列が入力される
      for(int i =0; i<N; i++){
      	  p[i] =sc.nextInt();        
      }
      
      //0番目とN番目を除いてループする
      //2番目に小さい数字をカウントしていく
      for(int i =1; i<N-1; i++){
        
      	if(p[i-1]<p[i] && p[i]<p[i+1]){
          count++;
        }
        else if(p[i-1]>p[i] && p[i]>p[i+1]){
          count++;
        }
      }
        System.out.println(count);
    }
}