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 (Clang 10.0.0) (judge verdict: Accepted)

#include <stdio.h>

int main(void)
{
	long long int num;
	long long int sum;
	long long int fizz;
	long long int buzz;
	long long int fizzbuzz;

	//Junretu Sum.
	scanf("%lld", &num);
	sum = (num * (2 + (num - 1))) / 2;

	//fizz sum.
	fizz = num / 3;
	fizz = (fizz * (2 * 3 + (fizz * 3 - 3))) / 2;

	//buzz sum.
	buzz = num / 5;
	buzz = (buzz * (2 * 5 + (buzz * 5 - 5))) / 2;

	//fizz buzz sum.
	fizzbuzz = num / 15;
	fizzbuzz = (fizzbuzz * (2 * 15 + (fizzbuzz * 15 - 15))) / 2;

	sum = sum - fizz - buzz + fizzbuzz;

	printf("%lld\n", sum);
	return (0);
}