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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;

#[rustfmt::skip]
fn main() { std::thread::Builder::new().stack_size(128 * 1024 * 1024).spawn(solve).unwrap().join().unwrap(); }

#[proconio::fastout]
fn solve() {
    input! {
        n: i64
    }
    println!(
        "{}",
        (1..=n)
            .map(|x| x * (x % 3 != 0) as i64 * (x % 5 != 0) as i64)
            .sum::<i64>()
    )
}