Problem:
Score : 200 points

Problem Statement

Let us define the FizzBuzz sequence a_1,a_2,... as follows:

If both 3 and 5 divides i, a_i=\mbox{FizzBuzz}.

If the above does not hold but 3 divides i, a_i=\mbox{Fizz}.

If none of the above holds but 5 divides i, a_i=\mbox{Buzz}.

If none of the above holds, a_i=i.

Find the sum of all numbers among the first N terms of the FizzBuzz sequence.

Constraints

1 \leq N \leq 10^6

Input

Input is given from Standard Input in the following format:

N

Output

Print the sum of all numbers among the first N terms of the FizzBuzz sequence.

Sample Input 1

15

Sample Output 1

60

The first 15 terms of the FizzBuzz sequence are:

1,2,\mbox{Fizz},4,\mbox{Buzz},\mbox{Fizz},7,8,\mbox{Fizz},\mbox{Buzz},11,\mbox{Fizz},13,14,\mbox{FizzBuzz}

Among them, numbers are 1,2,4,7,8,11,13,14, and the sum of them is 60.

Sample Input 2

1000000

Sample Output 2

266666333332

Watch out for overflow.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;

#define SZ(x) (int)(x.size())

using ll = long long;
using ld = long double;
using P = pair<int, int>;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
const ld eps = 1e-9;
const ll MOD = 1000000007;
const int INF = 1000000000;
const ll LINF = 1ll<<50;

template<typename T>
void printv(const vector<T>& s) {
  for(int i=0;i<(int)(s.size());++i) {
    cout << s[i];
    if(i == (int)(s.size())-1) cout << endl;
    else cout << " ";
  }
}

template<typename T1, typename T2>
ostream& operator<<(ostream &os, const pair<T1, T2> p) {
  os << p.first << ":" << p.second;
  return os;
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(10);

  ll n; cin >> n;
  ll su = 0;
  for(int i=1;i<=n;++i) {
    if(i%3 != 0 && i%5 != 0) {
      su += i;
    }
  }
  cout << su << endl;
  
}