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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

  public static void main(String[] args) {

    Scanner sc = new Scanner(System.in);

    int n = sc.nextInt();
    int[] p = new int[n];
    int ans = 0;

    for (int i = 0; i < n; i++) {
      p[i] = sc.nextInt();
    }

    for (int i = 0; i < (n - 2); i++) {
      if (p[i + 1] == med(p[i], p[i + 1], p[i + 2])) {
        ans++;
      }
    }

    System.out.println(ans);

  }

  // 3数の中央値を求めるメソッド
  static int med(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;
    }
  }

}