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 all(x) (x).begin(), (x).end()
#define sz(x) (int)x.size()
#define D(x) cerr << #x << " = " << (x) << '\n'
using ll = long long;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	ll n;
	cin >> n;
	auto sum_div = [&](ll n, ll k) {
		return k * (n / k * (n / k + 1)) / 2;
	};
	cout << sum_div(n, 1) - sum_div(n, 3) - sum_div(n, 5) + sum_div(n, 15) << '\n';
}