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)

#include <stdio.h>
#include <math.h>
int main(void) {
    int i, n;
    int sum = 0;
    while(1) {
        scanf("%d", &n);
        if(1 <= n && n <= 10^6) {
            break;
        }
        else {
            continue;
        }
    }
    for(i = 1; i <= n; i++) {
        if(i%3 == 0 && i%5 == 0) {
            continue;
        }
        else if(i%3 == 0) {
            continue;
        }
        else if(i%5 == 0) {
            continue;
        }
        else {
            sum = i + sum;
        }
    }
    printf("%d\n", sum);
    return 0;
}