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 Rust (1.15.1) (judge verdict: Accepted)

fn read<T: std::str::FromStr>() -> T {
  let mut s = String::new();
  std::io::stdin().read_line(&mut s).ok();
  s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
  read::<String>().split_whitespace()
    .map(|e| e.parse().ok().unwrap()).collect()
}

fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
  (0..n).map(|_| read_vec()).collect()
}

pub trait LexicalPermutation {
  /// Return `true` if the slice was permuted, `false` if it is already
  /// at the last ordered permutation.
  fn next_permutation(&mut self) -> bool;
}
impl<T> LexicalPermutation for [T] where T: Ord {
  /// Original author in Rust: Thomas Backman <serenity@exscape.org>
  fn next_permutation(&mut self) -> bool {
    // These cases only have 1 permutation each, so we can't do anything.
    if self.len() < 2 { return false; }

    // Step 1: Identify the longest, rightmost weakly decreasing part of the vector
    let mut i = self.len() - 1;
    while i > 0 && self[i-1] >= self[i] {
      i -= 1;
    }

    // If that is the entire vector, this is the last-ordered permutation.
    if i == 0 {
      return false;
    }

    // Step 2: Find the rightmost element larger than the pivot (i-1)
    let mut j = self.len() - 1;
    while j >= i && self[j] <= self[i-1]  {
      j -= 1;
    }

    // Step 3: Swap that element with the pivot
    self.swap(j, i-1);

    // Step 4: Reverse the (previously) weakly decreasing part
    self[i..].reverse();

    true
  }
}
fn main() {
  let n: usize = read();
  let p: Vec<usize> = read_vec();

  let mut count = 0;
  for i in 1..p.len()-1 {
    if (p[i-1] > p[i] && p[i] > p[i+1]) || (p[i-1] < p[i] && p[i] < p[i+1]) {
      count += 1;
    }
  }
  println!("{}", count);
}