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 vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vs = vector<string>;
using pll = pair<ll, ll>;
using vp = vector<pll>;
#define rep(i, n) for (ll i = 0; i < (n); i++)
#define repr(i, a, b) for (ll i = (a); i < (b); i++)
#define ALL(a) (a).begin(), (a).end()
#define SZ(x) ((ll)(x).size())
#define PRINT(V) for (auto v:(V)) {cerr << v << " ";} cerr << endl
#define dump(x) cerr << #x << " : " << (x) << endl
#define dump2(x, y) cerr << "(" << #x << ", " << #y << ") = (" << x << ", " << y << ")" << endl
inline ll GCD(ll a, ll b){return b ? GCD(b, a % b) : a;}
const ll MOD = 1000000007;
const ll INF = 100000000000000000LL;
/* END OF TEMPLATE */

int main(){
    cin.tie(0); ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    ll n;
    cin >> n;
    vll p(n);
    rep(i, n){
        cin >> p[i];
    }
    ll ans = 0;
    repr(i, 1, n-1){
        vll x(3);
        x[0] = p[i-1];
        x[1] = p[i];
        x[2] = p[i+1];
        sort(ALL(x));
        if(x[1] == p[i]){
            ans++;
        }
    }
    cout << ans << endl;
    return 0;
}