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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.*

fun main(args: Array<String>) {
    val argList = Scanner(System.`in`)


    val n = argList.nextLine().toInt()
    val p = argList.nextLine().map { it.toInt() }

    val executor = OrdinaryNumber()
    val result = executor.execute(n, p)
    println(result)
}

class OrdinaryNumber {
    fun execute(n: Int, p: List<Int>): Int {
        var count = 0

        val beginIndex = 1
        val beforeEndIndex = n - 1
        for (i in beginIndex..beforeEndIndex) {
            val targetList = listOf(p[i - 1], p[i], p[i + 1])
            val sorted = targetList.sorted()

            // すべて同じとき、2番目小さいに該当しない
            if (sorted.distinct().size == 1) continue

            if (p[i] == sorted[1]) {
                count++
            }
        }
        // 一番大きいとき　2番めのとき　3番目のとき
        // すべて等しいとき　2つ等しいとき
        return count
    }
}