Problem:
Score : 600 points

Problem Statement

In Dwango Co., Ltd., there is a content distribution system named 'Dwango Media Cluster', and it is called 'DMC' for short.

The name 'DMC' sounds cool for Niwango-kun, so he starts to define DMC-ness of a string.

Given a string S of length N and an integer k (k \geq 3),
he defines the k-DMC number of S as the number of triples (a, b, c) of integers that satisfy the following conditions:

0 \leq a < b < c \leq N - 1

S[a] = D

S[b] = M

S[c] = C

c-a < k

Here S[a] is the a-th character of the string S. Indexing is zero-based, that is, 0 \leq a \leq N - 1 holds.

For a string S and Q integers k_0, k_1, ..., k_{Q-1}, calculate the k_i-DMC number of S for each i (0 \leq i \leq Q-1).

Constraints

3 \leq N \leq 10^6

S consists of uppercase English letters

1 \leq Q \leq 75

3 \leq k_i \leq N

All numbers given in input are integers

Input

Input is given from Standard Input in the following format:

N
S
Q
k_{0} k_{1} ... k_{Q-1}

Output

Print Q lines.
The i-th line should contain the k_i-DMC number of the string S.

Sample Input 1

18
DWANGOMEDIACLUSTER
1
18

Sample Output 1

1

(a,b,c) = (0, 6, 11) satisfies the conditions.

Strangely, Dwango Media Cluster does not have so much DMC-ness by his definition.

Sample Input 2

18
DDDDDDMMMMMCCCCCCC
1
18

Sample Output 2

210

The number of triples can be calculated as 6\times 5\times 7.

Sample Input 3

54
DIALUPWIDEAREANETWORKGAMINGOPERATIONCORPORATIONLIMITED
3
20 30 40

Sample Output 3

0
1
2

(a, b, c) = (0, 23, 36), (8, 23, 36) satisfy the conditions except the last one, namely, c-a < k_i.

By the way, DWANGO is an acronym for "Dial-up Wide Area Network Gaming Operation".

Sample Output 4

30
DMCDMCDMCDMCDMCDMCDMCDMCDMCDMC
4
5 10 15 20

Sample Output 4

10
52
110
140

Solution in Haskell (GHC 7.10.3) (judge verdict: Time Limit Exceeded)

{-# LANGUAGE BangPatterns #-}
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE TupleSections #-}
import qualified Data.ByteString.Lazy.Char8 as B
import Data.Maybe
import Data.List
import qualified Data.Vector.Unboxed as VU
-- import qualified Data.Vector.Unboxed.Mutable as VUM
-- import Control.Monad
-- import Control.Monad.ST
-- import Debug.Trace

solve :: Int -> String -> Int -> [Int] -> [Int]
solve n s q ks = count
  where

  sv = VU.fromList s

  appD = mkCntVec n 'D' sv
  appC = mkCntVec n 'C' sv
  posM = [i | i <- [0..n-1], sv VU.! i == 'M']

  count = foldr1 (zipWith (+)) lst
    where
    (_, lst) = mapAccumL op (replicate q 0) (zip ((-1):posM) posM)
    op thisCnt (jm1,jm2) = (newCnt, newCnt)
      where
      newCnt = zipWith3 (\c i d -> c + (i - d)) thisCnt inc dec
      inc = map g ks
        where g k = sum $ map f [i | i <- [jm1+1..jm2-1], sv VU.! i == 'D']
                where f i | i+k <= jm2 = 0
                          | otherwise = (appC VU.! (min (i+k-1) (n-1)))
                                        - (appC VU.! jm2)
      dec = map g ks
        where g k = sum $ map f [i | i <- [jm1+1..jm2-1], sv VU.! i == 'C']
                where f i | jm1 < 0      = 0
                          | jm1 <= i-k+1 = 0
                          | i-k < 0      =  appD VU.! jm1
                          | otherwise    = (appD VU.! jm1) - (appD VU.! (i-k))

mkCntVec n c0 sv = VU.unfoldrN n f (0,0)
  where f (cnt, i) | sv VU.! i == c0 = Just (cnt+1, (cnt+1, i+1))
                   | otherwise       = Just (cnt, (cnt, i+1))

readBInt :: B.ByteString -> Int
readBInt = fst . fromJust . B.readInt

tmain :: B.ByteString -> [Int]
tmain cont =
  let remLines0 = map B.words (B.lines cont)
      [bs_n]:remLines1 = remLines0
      n = readBInt bs_n
      [bs_s]:remLines2 = remLines1
      s = B.unpack bs_s
      [bs_q]:remLines3 = remLines2
      q = readBInt bs_q
      line4:remLines4 = remLines3
      ks = map readBInt line4
  in solve n s q ks

outAnswer :: [Int] -> IO ()
outAnswer xs = putStr $ unlines $ map show xs

main :: IO ()
main = outAnswer . tmain =<< B.getContents