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: Accepted)

let () = Scanf.scanf "%d %d" @@ fun l r ->
  Printf.printf "%d\n" @@
  if
    l = 0 ||
    1 <= r / 673 - (l - 1) / 673 &&
    1 <= r / 3 - (l - 1) / 3
  then 0
  else
    Array.fold_left (Array.fold_left min) max_int @@
    Array.init (r - l + 1) @@ fun i ->
      Array.init (r - l + 1) @@ fun j ->
        let i = i + l in
        let j = j + l in
        if j <= i
        then max_int
        else (i * j) mod 2019