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)

{-# OPTIONS_GHC -O2 #-}
{-# LANGUAGE BangPatterns          #-}
{-# LANGUAGE CPP                   #-}
{-# LANGUAGE MultiParamTypeClasses #-}
{-# LANGUAGE OverloadedStrings     #-}
{-# LANGUAGE RecordWildCards       #-}
{-# LANGUAGE TupleSections         #-}
{-# LANGUAGE TypeFamilies          #-}

import           Control.Applicative
import           Control.Monad
import           Control.Monad.Primitive
import           Control.Monad.ST
import           Control.Monad.State.Strict
import           Data.Bool
import qualified Data.ByteString.Char8       as B
import qualified Data.ByteString.Unsafe      as B
import           Data.Char
import           Data.Function
import qualified Data.IntMap.Strict          as IM
import qualified Data.IntSet                 as IS
import           Data.List
import qualified Data.Map.Strict             as M
import           Data.Monoid
import           Data.Ord
import           Data.Primitive.MutVar
import qualified Data.Set                    as S
import           Data.Tuple
import qualified Data.Vector                 as V
import qualified Data.Vector.Generic         as G
import qualified Data.Vector.Generic.Mutable as GM
import qualified Data.Vector.Unboxed         as U
import qualified Data.Vector.Unboxed.Mutable as UM
import           Debug.Trace
import           Foreign                     hiding (void)
import           Unsafe.Coerce

main :: IO ()
main = do
    n <- readLn :: IO Int
    cs <- U.fromList <$> getLine
    q <- readLn :: IO Int
    ks <- map read.words <$> getLine
    mapM_ (print.solve cs) ks

solve :: U.Vector Char -> Int -> Int
solve cs k = fst $ U.foldl' step (U.foldl' step0 (0, emptyQ) pre) post
  where
    (pre, post) = U.splitAt k cs
    step0 (!res, !q) c
        | c == 'C' = (res + score q', q')
        | otherwise = (res, q')
      where
        q' = push c q

    step (!res, !q) c
        | c == 'C' = (res + score q', q')
        | otherwise = (res, q')
      where
        q' = push c $ pop q

data Q = Q !Int !Int !Int !String !String

emptyQ :: Q
emptyQ = Q 0 0 0 [] []

push :: Char -> Q -> Q
push 'D' (Q d m dm fs rs) = Q (d + 1) m dm fs ('D':rs)
push 'M' (Q d m dm fs rs) = Q d (m + 1) (dm + d) fs ('M':rs)
push c (Q d m dm fs rs)   = Q d m dm fs (c:rs)

pop :: Q -> Q
pop (Q d m dm ('D':fs) rs) = Q (d - 1) m (dm - m) fs rs
pop (Q d m dm ('M':fs) rs) = Q d (m - 1) dm fs rs
pop (Q d m dm (f:fs) rs)   = Q d m dm fs rs
pop (Q d m dm [] rs)       = pop $ Q d m dm (reverse rs) []

score :: Q -> Int
score (Q _ _ dm _ _) = dm