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;


//Optimisations

#pragma GCC target ("avx2")
#pragma GCC optimization ("unroll-loops")
#pragma GCC optimize("O2")


//shortcuts for functions

#define pb push_back
#define mp make_pair
#define ff first 
#define ss second
#define endl "\n"
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define th(n) cout<<n<<endl
#define gc getchar_unlocked
#define ms(s, n) memset(s, n, sizeof(s))
#define prec(n) fixed<<setprecision(n)
#define n_l '\n'


// make it python 

#define gcd __gcd
#define append push_back
#define str to_string
#define upper(s) transform(s.begin(),s.end(),s.begin(),::toupper)
#define lower(s) transform(s.begin(),s.end(),s.begin(),::tolower)
#define print(arr) for(auto el: arr) cout<<el<<" ";cout<<endl



// utility functions shortcuts

#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define sswap(a,b) {a=a^b;b=a^b;a=a^b;}
#define swap(a,b) {auto temp=a; a=b; b=temp;}
#define init(dp) memset(dp,-1,sizeof(dp));
#define set0(dp) memset(dp,0,sizeof(dp));
#define bits(x) __builtin_popcount(x)
#define SORT(v) sort(all(v))
#define endl "\n"
#define forr(i,n) for(ll i=0;i<n;i++)
#define formatrix(i,n) for(ll i=0;i<n;i++, cout<<"\n")
#define eof (scanf("%d" ,&n))!=EOF




// declaration shortcuts

#define vi vector<int>
#define vll vector<ll>
#define vvi vector<vector<int>>
#define vvl vector<vector<ll>> 
#define pll pair<ll,ll>
#define ppl pair<ll,pp>
#define ull unsigned long long
#define ll long long
#define mll map< ll, ll > 
#define sll set< ll > 
#define uni(v)   v.erase(unique(v.begin(),v.end()),v.end());
#define ini(a, v)       memset(a, v, sizeof(a))



// Constants

constexpr int dx[] = {-1, 0, 1, 0, 1, 1, -1, -1};
constexpr int dy[] = {0, -1, 0, 1, 1, -1, 1, -1};
constexpr ll INF = 1999999999999999997; 
constexpr int inf= INT_MAX;
constexpr int MAXSIZE = int(1e6)+5;
constexpr auto PI  = 3.14159265358979323846L;
constexpr auto oo  = numeric_limits<int>::max() / 2 - 2;
constexpr auto eps = 1e-6;
constexpr auto mod = 1000000007;
constexpr auto MOD = 1000000007;
constexpr auto MOD9 = 1000000009;

// Debugging 

// For reference: https://codeforces.com/blog/entry/65311

#define dbg(...) cout << "[" << #__VA_ARGS__ << "]: "; cout << to_string(__VA_ARGS__) << endl
template <typename T, size_t N> int SIZE(const T (&t)[N]){ return N; } template<typename T> int SIZE(const T &t){ return t.size(); } string to_string(string s, int x1=0, int x2=1e9){ return '"' + ((x1 < s.size()) ? s.substr(x1, x2-x1+1) : "") + '"'; } string to_string(const char* s) { return to_string((string) s); } string to_string(bool b) { return (b ? "true" : "false"); } string to_string(char c){ return string({c}); } template<size_t N> string to_string(bitset<N> &b, int x1=0, int x2=1e9){ string t = ""; for(int __iii__ = min(x1,SIZE(b)),  __jjj__ = min(x2, SIZE(b)-1); __iii__ <= __jjj__; ++__iii__){ t += b[__iii__] + '0'; } return '"' + t + '"'; } template <typename A, typename... C> string to_string(A (&v), int x1=0, int x2=1e9, C... coords); int l_v_l_v_l = 0, t_a_b_s = 0; template <typename A, typename B> string to_string(pair<A, B> &p) { l_v_l_v_l++; string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; l_v_l_v_l--; return res; } template <typename A, typename... C> string to_string(A (&v), int x1, int x2, C... coords) { int rnk = rank<A>::value; string tab(t_a_b_s, ' '); string res = ""; bool first = true; if(l_v_l_v_l == 0) res += n_l; res += tab + "["; x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v)); auto l = begin(v); advance(l, x1); auto r = l; advance(r, (x2-x1) + (x2 < SIZE(v))); for (auto e = l; e != r; e = next(e)) { if (!first) { res += ", "; } first = false; l_v_l_v_l++; if(e != l){ if(rnk > 1) { res += n_l; t_a_b_s = l_v_l_v_l; }; } else{ t_a_b_s = 0; } res += to_string(*e, coords...); l_v_l_v_l--; } res += "]"; if(l_v_l_v_l == 0) res += n_l; return res; } void dbgs(){;} template<typename Heads, typename... Tails> void dbgs(Heads H, Tails... T){ cout << to_string(H) << " | "; dbgs(T...); }

#define dbgm(...) cout << "[" << #__VA_ARGS__ << "]: "; dbgs(__VA_ARGS__); cout << endl;
#define n_l '\n'


int main()
{
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ll n;
  cin>>n;
  vll a(n);
  ll ans = 0;
  forr(i,n)
    cin>>a[i];
  for(int i = 1;i<n-1;i++)
  {
    if(a[i - 1] > a[i] and a[i+1] <= a[i])
      ans++;
    else if(a[i + 1] > a[i] and a[i- 1] <= a[i])
      ans++;
  }
  th(ans);
  // lvll 
}