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.PriorityQueue;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner std = new Scanner(System.in);
        int n = std.nextInt();
        int[] p = new int[n];
        for (int i = 0; i < n; i++) {
            p[i] = std.nextInt();
        }

        long count = 0;
        for (int i = 1; i < n - 1; i++) {
            PriorityQueue<Integer> pq = new PriorityQueue<>(3);
            int left = p[i - 1];
            int right = p[i + 1];
            int v = p[i];
            pq.add(left);
            pq.add(right);
            pq.add(v);
            pq.poll();
            if (v == pq.poll()) {
                count++;
            }
        }

        System.out.println(count);
    }
}