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<stdio.h>
int main(void)
{
    long int n;
    scanf("%ld",&n);
    long int a = n / 3;
    long int b = n / 5;
    long int c = n / 15;
    long int sum,sum3,sum5,sum15;
    sum = n*(n+1)/2;
    sum3 = 3*a*(a+1)/2;
    sum5 = 5*b*(b+1)/2;
    sum15 = 15*c*(c+1)/2;
    long int ans = sum -sum3 -sum5 + sum15;
    printf("%ld\n",ans);
    return 0;
}