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>
#define eps 1e-8
#define pr pair<int,int>
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int maxx = 1e5;

int main()
{
	int n;
	float a, b, c;
	ll sum = 0;
	while(cin >> n)
	{
		for(int i = 1; i <= n; i ++)
		{
			a = 1.0 * i / 3;
//			cout << i << endl;
			b = 1.0 * i / 5;
//			cout << i << '\n';
			c = 1.0 * i / 15;
			if(c == int(c))
			{
				continue;
			}
			else if(a == int(a)){
				continue;
			}
			else if(b == int(b)){
				continue;
			}
			else sum += i;
		}
		printf("%lld\n", sum);
		
	}
	return 0;
}