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;
#define ll long long
#define pb push_back
#define all(t) t.begin(), t.end()
#define inrange(i, a, b) (((i)>= min((a), (b))) && ((i) <= max((a), (b))))
typedef vector<ll> vi;
#define fi first
#define se second
const ll inf = 9e18;
const ll mod = 1e9 + 7;

void print(ll a[], ll n){for(ll i=0;i<n;i++){cout<<a[i]<<" ";}cout<<"\n";}
ll power(ll x, ll y, ll M = inf){
    ll ans = 1;
    x %= M;
    while(y){
        if(y&1)
            ans = (x * ans) % M;

        x = (x * x) % M;
        y >>= 1;
    }
    return ans;
}
ll modInverse(ll n) {return power(n, mod-2, mod);}
inline ll mul(ll a, ll b){ return (a * b) % mod; }
inline ll sub(ll a, ll b){ ll c = a - b; if(c < 0) c += mod; return c; }
inline ll add(ll a, ll b){ ll c = a + b; if(c > mod) c -= mod; return c; }
inline ll divi(ll a, ll b){ return mul(a, modInverse(b)); }

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);

	ll n;
	cin>>n;
	ll a[n];
	ll c=0;
	for(int i=0;i<n;i++){
		cin>>a[i];
	}
	for(int i=1;i<n-1;i++){
		if(a[i] > a[i-1] && a[i] < a[i+1]) c++;
		if(a[i] < a[i-1] && a[i] > a[i+1]) c++;
	}
	cout<<c;
	return 0;
}