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 std::io;

fn procon_read<T: std::str::FromStr>() ->  Vec<T> {
	let mut buf = String::new();
	io::stdin().read_line(&mut buf).expect("");
	let iter = buf.split_whitespace();
	let mut v: Vec<T> = Vec::new();
	for i in iter {
		v.push(i.parse().ok().unwrap());
	}
	v
}

fn main() {
	let n:u64 = procon_read()[0];
	println!("{}", n*(n+1)/2 - 3*(n/3)*(1 + n/3)/2 - 5*(n/5)*(1 + n/5)/2 + 15*(n/15)*(1 + n/15)/2);
}