Problem:
Score : 500 points

Problem Statement

Takahashi has a string S of length N consisting of digits from 0 through 9.

He loves the prime number P. He wants to know how many non-empty (contiguous) substrings of S - there are N \times (N + 1) / 2 of them - are divisible by P when regarded as integers written in base ten.

Here substrings starting with a 0 also count, and substrings originated from different positions in S are distinguished, even if they are equal as strings or integers.

Compute this count to help Takahashi.

Constraints

1 \leq N \leq 2 \times 10^5

S consists of digits.

|S| = N

2 \leq P \leq 10000

P is a prime number.

Input

Input is given from Standard Input in the following format:

N P
S

Output

Print the number of non-empty (contiguous) substrings of S that are divisible by P when regarded as an integer written in base ten.

Sample Input 1

4 3
3543

Sample Output 1

6

Here S = 3543. There are ten non-empty (contiguous) substrings of S:

3: divisible by 3.

35: not divisible by 3.

354: divisible by 3.

3543: divisible by 3.

5: not divisible by 3.

54: divisible by 3.

543: divisible by 3.

4: not divisible by 3.

43: not divisible by 3.

3: divisible by 3.

Six of these are divisible by 3, so print 6.

Sample Input 2

4 2
2020

Sample Output 2

10

Here S = 2020. There are ten non-empty (contiguous) substrings of S, all of which are divisible by 2, so print 10.

Note that substrings beginning with a 0 also count.

Sample Input 3

20 11
33883322005544116655

Sample Output 3

68

Solution in Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

import Data.Maybe
import Data.Char
import Data.Word
import qualified Data.IntMap.Strict as M
import qualified Data.ByteString.Char8 as C

parseInt :: C.ByteString -> Int
parseInt = fst . fromJust . C.readInt

suffixSums :: Int -> C.ByteString -> M.IntMap Int
suffixSums p s = snd $ C.foldr' acc ((1, 0), M.fromList [(i, 0) | i <- [0 .. p-1]]) s
  where
  acc c ((x, y), mp) = ((x', y'), M.adjust (\o -> o + 1) y' mp)
    where
    x' = 10 * x `mod` p
    y' = (x * digitToInt c + y) `mod` p

fpm :: Int -> Int -> Int -> Int
fpm p a e
  | e == 0 = 1
  | otherwise = (if odd e then a else 1) * aa `mod` p * aa `mod` p
  where
  aa = fpm p a $ e `div` 2

solve :: Int -> C.ByteString -> Word64
solve p s
  | n == 1 = if parseInt s `mod` p == 0 then 1 else 0
  | otherwise = cnt + solve p x + solve p y
  where
  n = C.length s
  (x, y) = C.splitAt (n `div` 2) s
  mp = suffixSums p x
  pp = fpm p 10 (p - 2)
  cnt = snd $ C.foldl' acc ((0, 1), 0) y
  acc ((val, pow), total) c =
    let val' = (val * 10 + digitToInt c) `mod` p
        pow' = pow * pp `mod` p
        o = (p - val') * pow' `mod` p
    in ((val', pow'), total + (fromIntegral $ fromJust $ M.lookup o mp))

main :: IO ()
main = do
  [_, p] <- map parseInt . C.words <$> C.getLine
  s <- C.getLine
  putStrLn $ show $ solve p s