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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// #include <ext/pb_ds/detail/standard_policies.hpp>
using namespace std;
using namespace __gnu_pbds;
// const ll RANDOM = chrono::high_resolution_clock::now().time_since_epoch().count();
// gp_hash_table<int, int> table;

#define ll long long int
#define vii vector< ll >
#define pb push_back
#define pii pair<ll,ll>
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define ins insert
#define ld long double
#define lp(l,r) for(i=l;i<=r;++i)

#define lpj(l,r) for(j=l;j<=r;++j)
#define lpr(l,r) for(i=r;i>=l;--i)
#define lprj(l,r) for(j=r;j>=l;--j)

#define swap(a,b){ auto _t=a;a=b,b=_t;}
#define flush() cout.flush()
#define popcount __builtin_popcount



// typedef __int128 bign;


#define TRACE

#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
	cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
	const char* comma = strchr(names + 1, ',');cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define trace(...)
#endif

#define MD          1000000007
#define M 			100011
#define LN 			14
#define fftmod      998244353
#define EPS         (double)1e-10

ll md=MD;
ll inf=1e18;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); 


inline ll exp(ll a,ll b){ll r=1LL;while(b>0){if(b&1){r=r*(a%md);r=(r+md)%md;}b/=2;a=(a%md)*(a%md);a=(a+md)%md;}return (r+md)%md;}
ll gcd(ll a,ll b){if(b==0)return a;if(a==0)return b;return gcd(b,a%b);}
inline ll poww(ll a,ll b){ll r=1LL;while(b>0){if(b&1)r=r*a;b/=2;a=a*a;}return (ll)r;}

using cd=complex<double>;
const double PI=acos(-1);

inline ll rand(ll l,ll r){
	return l+rng()%(r-l+1);
}

ll a[M];
int main(){


    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);

   // ifstream fin("input.txt");ofstream fout("outut.txt");
    ll t,i,j,m,n,s,z,k,p,T;string s1,s2;ll q;ll x,y;
    cin>>n;
    lp(1,n)cin>>a[i];
    p=0;
    for(i=2;i+1<=n;++i){
        vii v;
        v.pb(a[i-1]);v.pb(a[i]);v.pb(a[i+1]);
        sort(all(v));
        if(a[i]==v[1])++p;
    }cout<<p<<"\n";







    return 0;
}