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 OCaml (4.02.3) (judge verdict: Time Limit Exceeded)

let modulo = 2019
let ( *% ) a b = (a * b) mod modulo
               
let () =
  let (l, r) = Scanf.sscanf (input_line stdin) "%d %d" (fun a b -> (a, b)) in
  let rec loop_l : int -> int = fun i ->
    if i > r then 2019 else
      let rec loop_r : int -> int = fun j ->
        if j > r then 2019
        else min (i *% j) (loop_r (j + 1)) in
      min (loop_r (i + 1)) (loop_l (i + 1))
  in
  let ans = if l - r > 2019 then 0
            else loop_l l in
  print_int ans;
  print_newline ()