Problem:
Score : 400 points

Problem Statement

Given is a string S consisting of digits from 1 through 9.

Find the number of pairs of integers (i,j) (1 ≤ i ≤ j ≤ |S|) that satisfy the following condition:

Condition: In base ten, the i-th through j-th characters of S form an integer that is a multiple of 2019.

Constraints

1 ≤ |S| ≤ 200000

S is a string consisting of digits from 1 through 9.

Input

Input is given from Standard Input in the following format:

S

Output

Print the number of pairs of integers (i,j) (1 ≤ i ≤ j ≤ |S|) that satisfy the condition.

Sample Input 1

1817181712114

Sample Output 1

3

Three pairs - (1,5), (5,9), and (9,13) - satisfy the condition.

Sample Input 2

14282668646

Sample Output 2

2

Sample Input 3

2119

Sample Output 3

0

No pairs satisfy the condition.

Solution in Haskell (GHC 8.8.3) (judge verdict: Accepted)

{-# OPTIONS_GHC -O2 #-}
{-# LANGUAGE OverloadedStrings #-}
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE LambdaCase #-}
{-# LANGUAGE BangPatterns #-}
module Main where

import           Data.List
import           Control.Monad
import           Debug.Trace
import qualified Data.ByteString.Char8         as BC
import           Data.Maybe                     ( fromJust, fromMaybe )
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.IO as IOA
import qualified Data.Array.MArray as MA
import Control.Monad.ST
import System.IO (stdin)
import Control.Arrow
import qualified Data.Vector.Unboxed as UV
import qualified Data.Vector.Generic as GV



readTuple2 :: IO (Int, Int)
readTuple2 = do
    [a, b] <- readInts
    return (a, b)

readTuple3 :: IO (Int, Int, Int)
readTuple3 = do
    [a, b, c] <- readInts
    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 0 = 1
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

chooseTwo :: Int -> Int
chooseTwo 0 = 0
chooseTwo 1 = 0
chooseTwo a = a * (a - 1) `div` 2

solve :: BC.ByteString -> Int
solve s = sum $ chooseTwo <$> intCount where
    ints = (BC.unpack . BC.reverse) s
    intList :: [ModNum]
    !intList = zipWith (\time base -> base * pow 10 time)  [0..] $ fromIntegral . digitToInt <$> ints
    sumacc = scanl' (+) 0 intList
    intCount :: [Int]
    intCount = fmap snd $ M.toList $ foldl' (\m n -> M.alter (Just . maybe 1 (+1)) n m) M.empty sumacc

main = do
    n <- BC.getLine
    print $ solve n