Problem:
Score : 300 points

Problem Statement

You are given two non-negative integers L and R.
We will choose two integers i and j such that L \leq i < j \leq R.
Find the minimum possible value of (i \times j) \mbox{ mod } 2019.

Constraints

All values in input are integers.

0 \leq L < R \leq 2 \times 10^9

Input

Input is given from Standard Input in the following format:

L R

Output

Print the minimum possible value of (i \times j) \mbox{ mod } 2019 when i and j are chosen under the given condition.

Sample Input 1

2020 2040

Sample Output 1

2

When (i, j) = (2020, 2021), (i \times j) \mbox{ mod } 2019 = 2.

Sample Input 2

4 5

Sample Output 2

20

We have only one choice: (i, j) = (4, 5).

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

// Original: https://github.com/tanakh/competitive-rs
#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, [ $t:tt ]) => {
        {
            let len = read_value!($next, usize);
            (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
        }
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

#[allow(dead_code)]
fn chmin<T>(x: &mut T, y: T) -> bool
where
    T: PartialOrd + Copy,
{
    *x > y && {
        *x = y;
        true
    }
}

#[allow(dead_code)]
fn chmax<T>(x: &mut T, y: T) -> bool
where
    T: PartialOrd + Copy,
{
    *x < y && {
        *x = y;
        true
    }
}

#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, VecDeque};

fn main() {
    input!(l: i64, r: i64);
    let m = 2019 as i64;

    let mut vs = BTreeSet::new();
    for left in l..min(r + 1, l + 2019) {
        let a = left % m;
        vs.insert(a);
    }

    let mut right = r;
    let r_min = max(0, r - 2019);
    while right > max(l, r_min) {
        let a = right % m;
        vs.insert(a);
        right -= 1;
    }
    let mut cnt = 0;
    let mut ans = 1;
    for i in vs.iter() {
        cnt += 1;
        let a = i;
        ans *= *a;
        if cnt >= 2 {
            break;
        }
    }
    if cnt < 2 {
        let mut it = vs.iter();
        let a = it.next().unwrap();
        ans *= *a;
    }

    // for i in vs.iter() {
    //     println!("{}", i);
    // }
    println!("{}", ans);
}