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 <bits/stdc++.h>
using namespace std;
using ll = long long;
using Graph = vector<vector<int>>;

const ll LINF = 1e18;
const int INF = 1e9;
const ll MOD = 1000000007;

template<class T> inline bool chmin(T& a, T b){
    if(a > b){
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b){
    if(a < b){
        a = b;
        return true;
    }
    return false;
}
vector<bool> seen;
void dfs(const Graph &G, int v){
    seen[v] = true; // vを訪問済みにする
    
    for(auto next_v: G[v]){
        if(seen[next_v]) continue; // next_vが探索済みならスルー
        dfs(G, next_v); // 再帰的に探索
    }
}

int main(){

    int n;
    cin >> n;
    vector<int> v(n);
    for(int i = 0; i < n; i++){
        cin >> v[i];
    }
    int cnt = 0;
    for(int i = 0; i < n-2; i++){
        bool flg = false;
        if((v[i] < v[i+1]) && (v[i+1] < v[i+2])) flg = true;
        if((v[i] > v[i+1]) && (v[i+1] > v[i+2])) flg = true;
        if(flg) cnt++;
    }
    cout << cnt << endl;
    return 0;
}