#include <iostream>
using namespace std;

int absfunc(int a) {
  if (a > 0) return a;
  else return -a;
}

int main() {
  int N;
  cin >> N;
  pair<int,int> arr[1001];
  for (int i=0; i<N; i++) {
    int A,B;
    cin >> A >> B;
    arr[i] = {A,B};
  }

  int ans = 0;
  for (int i=0; i<N-1; i++) for (int j=i+1; j<N; j++) {
    int deltaX = arr[j].first - arr[i].first;
    int deltaY = arr[j].second - arr[i].second;
    if (absfunc(deltaX) >= absfunc(deltaY)) ans++;
  }

  cout << ans;
}
#include<bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp> //required
#include <ext/pb_ds/tree_policy.hpp> //required

using namespace __gnu_pbds; 
using namespace std;

template <typename T> using ordered_set =  tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; 
// find_by_order(k)  returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;
typedef long long  ll;
typedef vector<ll> VL;
typedef vector<int> VI;
typedef pair<ll,ll> PLL;
typedef pair<int,int> PII;
#define pb push_back

#define F first
#define S second
#define SZ(a) int((a).size())
#define ALL(a) a.begin(),a.end()
#define fr(i,x,y) for(int i=x;i<y;i++)
#define frr(i,x,y) for(int i=x-1;i>=y;i--)
#define here cout<<"here:\n"
#define inf 1e18+1
const long double PI = acos(-1.0L);
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
const int mod=1000000007;
//const int mod=998244353;
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
ll power(ll a,ll b){ll ans=1;while(b!=0){if(b&1){ans*=a;}a=a*a;b>>=1;}return ans;}
ll powerm(ll a,ll b){ll ans=1;while(b!=0){if(b&1){ans*=a;ans%=mod;}a=a*a;a%=mod;b>>=1;}return ans%mod;}
VL dx={1,0,-1,0};
VL dy={0,1,0,-1};
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();

void solve()
{
    ll n;
    cin>>n;
    VL x(n),y(n);
    fr(i,0,n)
    {
        cin>>x[i]>>y[i];
    }
    ll ct=0;
    fr(i,0,n)
    {
        fr(j,i+1,n)
        {
            double res=(double)(y[j]-y[i])/(double)(x[j]-x[i]);
            if(res>=-1.00&&res<=1.00)
            {
                ct++;
            }
        }
    }
    cout<<ct<<endl;
    return;
}
int main()
{
    IOS;
    ll t=1,pp;
    //cin>>t;
    pp=t;
    while(t--)
    {
        //cout<<"Case #"<<pp-t<<": ";
        solve();
    }
    return 0;
}
/* stuff you should look for
        * int overflow, array bounds
        * special cases (n=1?)
        * do smth instead of nothing and stay organized
        * WRITE STUFF DOWN
        * BE CAREFUL REGARDING THE DEFAULT VALUES IN segement trees etc
        * BE very careful in int vs long long vs unsigned long long 
*/
/*
recursion - matrix exponential

*/
// BITMASK:
// 1)When some constrall is of the order of 15-20, think of bitmask DP.
// 2)When some constrall is around 40, try out meet in the middle
// 3) See Strings,palindromes,prefix,suffix etc -> KMP,Z algorithmQ