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 Haskell (GHC 7.10.3) (judge verdict: Accepted)

{-# LANGUAGE OverloadedStrings #-}
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE LambdaCase #-}
module Main where

import           Data.List
import           Control.Monad
import           Debug.Trace
import qualified Data.ByteString.Char8         as BC
import           Data.Maybe                     ( fromJust )
import qualified Data.Set                      as S
import qualified Data.IntMap.Strict            as IM
import qualified Data.IntMap.Lazy            as IML
import qualified Data.Map.Strict               as M
import           Data.Char
import qualified Data.Array.IArray as A
import qualified Data.Array.Unboxed as UA
import Data.Bits
import Control.Monad.State.Strict
import qualified Data.Array.ST as STA
import qualified Data.Array.MArray as MA
import Control.Monad.ST



readTuple2 :: IO (Int, Int)
readTuple2 = do
    [a, b] <- map (fst . fromJust . BC.readInt) . BC.words <$> BC.getLine
    return (a, b)

readTuple3 :: IO (Int, Int, Int)
readTuple3 = do
    [a, b, c] <- map (fst . fromJust . BC.readInt) . BC.words <$> BC.getLine
    return (a, b, c)

readInt :: IO Int
readInt = fst . fromJust . BC.readInt <$> BC.getLine

readInts :: IO [Int]
readInts = map (fst . fromJust . BC.readInt) . BC.words <$> BC.getLine

split :: (a -> Bool) -> [a] -> [[a]]
split f [] = []
split f as =
    let (fo, la) = span f as
        las      = case la of
            [] -> []
            xs -> split f $ tail xs
    in  case fo of
            [] -> las
            xs -> xs : las

modN = 2019

pow :: Num a => a -> Int -> a
pow base 1 = base
pow base time
    | odd time = let 
        a = pow base (div time 2)
        in
        a * a * base
    | otherwise = let
        a = pow base (div time 2)
        in
        a * a

inv :: ModNum -> ModNum
inv a = pow a $ modN - 2

newtype ModNum = ModNum {getInt :: Int} deriving (Show, Eq, Ord)
instance Num ModNum where
    (+) (ModNum a) (ModNum b) = ModNum $ mod (a + b) modN
    (-) (ModNum a) (ModNum b)
            | a < b = (ModNum $ a + modN) - ModNum b
            | otherwise = ModNum $ mod (a - b) modN
    (*) (ModNum a) (ModNum b) = ModNum $ mod (a * b) modN
    abs (ModNum a)
        | a < 0 = abs . ModNum $ a + modN
        | otherwise = ModNum a
    signum (ModNum a) = ModNum $ signum a
    fromInteger i = abs . ModNum . fromIntegral $ mod i (fromIntegral modN)

mDiv :: ModNum -> ModNum -> ModNum
mDiv a b = a * inv b

comb :: Int -> [a] ->[[a]]
comb 0 xs = [[]]
comb _ [] = []
comb n (x:xs) = [x:y | y <- comb (n-1) xs] ++ comb n xs

main :: IO ()
main = do
    (l, r) <- readTuple2
    if r - l > 2019 then print 0 else print $ getInt $ minimum $ do
        [a, b] <- comb 2 $ fromIntegral <$> [l..r]
        pure $ a * b