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: Wrong Answer)

#pragma GCC optimize("O3")
#pragma GCC target("avx")
#include <stdio.h>
#include <stdlib.h>
#define min(a, b) ((a) < (b) ? (a) : (b))
#define max(a, b) ((a) > (b) ? (a) : (b))
#define abs(a) ((a) > 0 ? (a) : (-1 * (a)))
#define rep(i, n) for ((i) = 0; (i) < (n); (i)++)
#define rep2(i, s, e) for ((i) = (s); (i) < (e); (i)++)
#define LIM (1000000007)

int main(void)
{

    int n, i;
    long long s = 0;
    scanf("%d", &n);
    rep2(i, 1, n + 1)
    {
        if (i % 3 != 0 && i % 5 != 0)
            s += i;
    }
    printf("%d", s);

    return 0;
}