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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		// 整数の入力
		int n = sc.nextInt();
		// スペース区切りの整数の入力
        int[] p = new int[n];
        for(int i=0;i<n;i++){
          p[i]=sc.nextInt();
        }
        
        int count=0;
        for(int i=1;i<n-1;i++){
         if(p[i]==mid(p[i-1],p[i],p[i+1]))
             count++;
        }
		// 出力
		System.out.println(count);
	}
  public static int mid(int a,int b ,int c){
    if(a >= b) {
    		if(b >= c) {
    			return b;
    		}
    		else if(a <= c) {
    			return a;
    		}
    		else {
    			return c;
    		}
    	}
    	else if(a > c) {
    		return a;
    	}
    	else if(b > c) {
    		return c;
    	}
    	else {
    		return b;
    	}
  }
    
}