#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define int long long
using namespace std;

using ll = long long;
using ld = long double;
struct poi{
    ll x,y;
    poi(){
        x=0,y=0;
    }
    poi(ll xx,ll yy){
        x=xx,y=yy;
    }
};
poi operator-(poi a,poi b){return poi(a.x-b.x,a.y-b.y);};
ll cro(poi a,poi b){return a.x*b.y-b.x*a.y;}
//ToLeft测试，判断p是否在ab左边（逆时针）,1左,-1右,0在线上
int toleft(poi a,poi b,poi p){
    ll ans=cro((b-a) , (p-b));
    if(ans==0)return 0;
    if(ans>0)return 1;
    return -1;
}

void solve(){
    int n;
    cin>>n;
    ll ans=0;
    vector<poi> p(5*n+5);
    for(int i=0;i<n;i++){
        cin>>p[i].x>>p[i].y;
        p[i+3*n]=p[i+4*n]=p[i+2*n]=p[i+n]=p[i];
    }
    //p.push_back(p[0]);
    for(int i=n*2;i<n*3;i++){
        int j=i,k=i;
        while(p[k-1].y==p[k].y)k--;
        while(p[j+1].y==p[j].y)j++;
        if(toleft(p[k-1],p[j+1],p[i])==-1&&toleft(p[k-1],p[j+1],p[i])==-1&&p[j+1].y>=p[i].y&&p[k-1].y>=p[i].y)
        ans++;
        //cout<<i-n<<' '<<j-n<<' '<<k-n<<' '<<p[j+1].y<<' '<<p[k-1].y<<' '<<p[i].y<<endl;
        i=j;
    }
    cout<<ans;
}

signed main(){
    IOS;
    int t = 1;
//        cin >> t;
    while (t -- ){
        solve();
    }
    return 0;
}