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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;

public class OrdinaryNumber {
    public static void Main(string[] argc){
        int length = int.Parse(Console.ReadLine());
        string[] input = Console.ReadLine().Split(' ');
        int[] arr = input.Select(x => int.Parse(x) ).ToArray();

        int count = CountSecondarySmall(arr, length);
        Console.WriteLine(count);
    }

    public static int CountSecondarySmall(int[] numArr, int length){
        int count=0;
        for(int i=2; i<length; i++){
            int first = numArr[i-2];
            int second = numArr[i-1];
            int third = numArr[i];
            bool isSecondIsCenter = (first < second && second < third ) || (third < second && second < first);
            if (isSecondIsCenter) count++;
        }
        return count;
    }
}