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: Accepted)

import           Control.Monad
import qualified Data.ByteString.Char8       as BS
import           Data.Char
import           Data.Maybe
import qualified Data.Vector.Unboxed.Mutable as VM

readInt = fst . fromJust . BS.readInt
readIntList = map readInt . BS.words
getInt = readInt <$> BS.getLine
getIntList = readIntList <$> BS.getLine

main = do
  [n, p] <- getIntList
  s <- BS.getLine

  v <- VM.new p
  VM.set v (0::Int)
  VM.write v 0 1 -- U0 `mod` p == 0                                                                                       
  let solve25 i a | i > n = print a
                  | otherwise = do
                      let c = BS.index s (n-i)
                          di = ord c - ord '0'
                          x = if di `mod` p == 0 then n - i + 1 else 0
                      solve25 (i+1) (a+x)

      solve i a md m10 | i > n = print a
                       | otherwise = do
                           let c = BS.index s (n-i)
                               di = ord c - ord '0'
                               md' = (md + di * m10) `mod` p
                               m10' = (m10 * 10) `mod` p
                           y <- VM.read v md'
                           VM.write v md' (y+1)
                           solve (i+1) (a+y) md' m10'

  if 10 `mod` p == 0
    then solve25 1 0
    else solve 1 0 0 1