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;
typedef long long ll;
typedef pair<ll, ll> P;
#define DUMP(x) cout << #x << " = " << (x) << endl;
#define FOR(i, m, n) for (ll i = m; i < n; i++)
#define IFOR(i, m, n) for (ll i = n - 1; i >= m; i--)
#define REP(i, n) FOR(i, 0, n)
#define IREP(i, n) IFOR(i, 0, n)
#define FOREACH(x, a) for (auto&(x) : (a))
#define ALL(v) (v).begin(), (v).end()
#define SZ(x) ll(x.size())

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