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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>
#include <fstream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <numeric>
#include <functional>
#include <map>
#define REP(i,n) for(int (i)=0; (i)<(n); (i)++)
using namespace std;
using ll = long long;
const ll MOD = 1e9+7;

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  int n;
  cin >> n;
  int c, o1, o2;
  int ans = 0;
  cin >> o2;
  cin >> o1;
  for(int i = 2; i < n; i++){
    cin >> c;
    if(max(o1, max(o2, c)) != o1 && min(o1, min(o2, c)) != o1 ){
      //cout << o1 << endl;
      ans++;
    }
    o2 = o1;
    o1 = c;
  }
  cout << ans << endl;
  return 0;
}