Problem:
Score : 300 points

Problem Statement

You are given four integers A, B, C, and D. Find the number of integers between A and B (inclusive) that can be evenly divided by neither C nor D.

Constraints

1\leq A\leq B\leq 10^{18}

1\leq C,D\leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C D

Output

Print the number of integers between A and B (inclusive) that can be evenly divided by neither C nor D.

Sample Input 1

4 9 2 3

Sample Output 1

2

5 and 7 satisfy the condition.

Sample Input 2

10 40 6 8

Sample Output 2

23

Sample Input 3

314159265358979323 846264338327950288 419716939 937510582

Sample Output 3

532105071133627368

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::*;
use std::str::FromStr;

fn parse_line<T: FromStr>() -> T {
    let mut s = String::new();
    stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn parse_quadraple<T: FromStr, U: FromStr, V: FromStr, W: FromStr>() -> (T, U, V, W) {
    let line = parse_line::<String>();
    let mut iter = line.split_whitespace();
    (
        iter.next().unwrap().parse().ok().unwrap(),
        iter.next().unwrap().parse().ok().unwrap(),
        iter.next().unwrap().parse().ok().unwrap(),
        iter.next().unwrap().parse().ok().unwrap(),
    )
}

fn println<T: ToString>(s: T) {
    println!("{}", s.to_string());
}

// ------------------------------------------------------------------------

fn gcd(x: u64, y: u64) -> u64 {
    let mut x = x;
    let mut y = y;
    while y != 0 {
        let t = y;
        y = x % y;
        x = t;
    }
    x
}

// ------------------------------------------------------------------------

fn count_dividable(min_n: u64, max_n: u64, divisor: u64) -> u64 {
    let first_valid = if min_n % divisor == 0 { min_n } else { min_n + divisor - min_n % divisor };

    1 + (max_n - first_valid) / divisor
}

fn main() {
    let (a, b, c, d) = parse_quadraple::<u64, u64, u64, u64>();

    let lcm = c * d / gcd(c, d);

    println(
        b + 1 - a - (count_dividable(a, b, c) + count_dividable(a, b, d) - count_dividable(a, b, lcm))
    );
}