Problem:
Score : 400 points

Problem Statement

At an arcade, Takahashi is playing a game called RPS Battle, which is played as follows:

The player plays N rounds of Rock Paper Scissors against the machine. (See Notes for the description of Rock Paper Scissors. A draw also counts as a round.)

Each time the player wins a round, depending on which hand he/she uses, he/she earns the following score (no points for a draw or a loss):

R points for winning with Rock;

S points for winning with Scissors;

P points for winning with Paper.

However, in the i-th round, the player cannot use the hand he/she used in the (i-K)-th round. (In the first K rounds, the player can use any hand.)

Before the start of the game, the machine decides the hand it will play in each round. With supernatural power, Takahashi managed to read all of those hands.

The information Takahashi obtained is given as a string T. If the i-th character of T (1 \leq i \leq N) is r, the machine will play Rock in the i-th round. Similarly, p and s stand for Paper and Scissors, respectively.

What is the maximum total score earned in the game by adequately choosing the hand to play in each round?

Notes

In this problem, Rock Paper Scissors can be thought of as a two-player game, in which each player simultaneously forms Rock, Paper, or Scissors with a hand.

If a player chooses Rock and the other chooses Scissors, the player choosing Rock wins;

if a player chooses Scissors and the other chooses Paper, the player choosing Scissors wins;

if a player chooses Paper and the other chooses Rock, the player choosing Paper wins;

if both players play the same hand, it is a draw.

Constraints

2 \leq N \leq 10^5

1 \leq K \leq N-1

1 \leq R,S,P \leq 10^4

N,K,R,S, and P are all integers.

|T| = N

T consists of r, p, and s.

Input

Input is given from Standard Input in the following format:

N K
R S P
T

Output

Print the maximum total score earned in the game.

Sample Input 1

5 2
8 7 6
rsrpr

Sample Output 1

27

The machine will play {Rock, Scissors, Rock, Paper, Rock}.

We can, for example, play {Paper, Rock, Rock, Scissors, Paper} against it to earn 27 points.
We cannot earn more points, so the answer is 27.

Sample Input 2

7 1
100 10 1
ssssppr

Sample Output 2

211

Sample Input 3

30 5
325 234 123
rspsspspsrpspsppprpsprpssprpsr

Sample Output 3

4996

Solution in Haskell (GHC 7.10.3) (judge verdict: Accepted)

--{-# LANGUAGE BangPatterns #-}
--{-# LANGUAGE ScopedTypeVariables #-}
--{-# LANGUAGE FlexibleContexts #-}
--{-# LANGUAGE ViewPatterns #-}
--{-# LANGUAGE OverloadedStrings #-}
--{-# LANGUAGE Strict #-}
--{-# LANGUAGE NumericUnderscores #-}
--{-# LANGUAGE BlockArguments #-}
--{-# LANGUAGE MultiWayIf #-}

import qualified Data.Vector as V
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import qualified Data.ByteString.Char8 as BC
import qualified Data.Map as MP
import qualified Data.Map.Strict as MPS
import qualified Data.Set as ST
import qualified Data.Sequence as SQ
-- import Data.Array
import Data.Array.Unboxed
-- import Data.Array.ST
-- import Data.Array.IO
import Data.Maybe
import Data.Ord
import Data.Ratio
import Data.Char
import Data.Bits
import Data.List
import Data.Tuple
import Data.Functor
import Data.IORef
import Data.STRef
import Control.Applicative
import Control.Monad
import Control.Monad.ST
import Debug.Trace
-- import qualified Data.Vector.Algorithms.Merge as VAM
-- import qualified Data.Heap as HP
-- import Numeric.Extra
-- import Data.Tuple.Extra
-- import Data.List.Extra

--------------------------------------------------------------------------
main = do
    [n,k] <- sLineToIntL
    [r,s,p] <- sLineToIntL
    ts <- strBS
    let
        solve = foldl' step 0 [0..k-1]
            where
                step acc i = acc + (snd $ foldl' inner ('x',0) [i,i+k..n-1])
                inner (prevHand,a) j
                    | prevHand == h = ('x',a)
                    | otherwise = (h,a+point)
                    where
                        h = BC.index ts j
                        point =
                            case h of
                                'r' -> p
                                's' -> r
                                'p' -> s
                            
    print solve
--------------------------------------------------------------------------

-- Input Util
int :: IO Int
int = BC.getLine >>= return . bsToInt

double :: IO Double
double = BC.getLine >>= return . bsToDouble

str :: IO String
str = BC.getLine >>= return . BC.unpack

strBS :: IO BC.ByteString
strBS = BC.getLine

-- convert
bsToInt :: BC.ByteString -> Int
bsToInt = fst . fromJust . BC.readInt

bsToDouble :: BC.ByteString -> Double
bsToDouble = read . BC.unpack

bsToIntTuple :: BC.ByteString -> (Int, Int)
bsToIntTuple bs = (a,b)
    where
        Just (a,bs') = parseInt bs
        Just (b,_) = parseInt bs'

bsToIntTriple :: BC.ByteString -> (Int, Int, Int)
bsToIntTriple bs = (a,b,c)
    where
        Just (a,bs') = parseInt bs
        Just (b,bs'') = parseInt bs'
        Just (c,_) = parseInt bs''

parseInt :: BC.ByteString -> Maybe (Int, BC.ByteString)
parseInt = BC.readInt . BC.dropWhile isSpace

-- single line
fromSLineL :: (BC.ByteString -> a) -> IO [a]
fromSLineL f = strBS >>= return . map f . BC.words

-- from single line to List / VU.Vector 
sLineToIntL :: IO [Int]
sLineToIntL = fromSLineL bsToInt

sLineToDoubleL :: IO [Double]
sLineToDoubleL = fromSLineL bsToDouble

sLineToStrL :: IO [String]
sLineToStrL = fromSLineL BC.unpack

sLineToStrBSL :: IO [BC.ByteString]
sLineToStrBSL = fromSLineL id 

sLineToIntV :: Int -> IO (VU.Vector Int)
sLineToIntV n = BC.getLine >>= return . VU.unfoldrN n parseInt

sLineToDoubleV :: IO (VU.Vector Double)
sLineToDoubleV = BC.getLine >>= return . VU.fromList . map bsToDouble . BC.words

-- multiple lines
fromMLinesL :: Int -> (BC.ByteString -> a) -> IO [a]
fromMLinesL n f = replicateM n (strBS >>= return . f)

fromMLinesV :: VU.Unbox a => Int -> (BC.ByteString -> a) -> IO (VU.Vector a)
fromMLinesV n f = VU.replicateM n (strBS >>= return . f)

-- from multiple lines to List / VU.Vector
mLinesToIntL :: Int -> IO [Int]
mLinesToIntL n = fromMLinesL n bsToInt

mLinesToDoubleL :: Int -> IO [Double]
mLinesToDoubleL n = fromMLinesL n bsToDouble

mLinesToStrL :: Int -> IO [String]
mLinesToStrL n = fromMLinesL n BC.unpack

mLinesToStrBSL :: Int -> IO [BC.ByteString]
mLinesToStrBSL n = fromMLinesL n id

mLinesToTupleL :: Int -> IO [(Int, Int)]
mLinesToTupleL n = fromMLinesL n bsToIntTuple

mLinesToTripleL :: Int -> IO [(Int, Int, Int)]
mLinesToTripleL n = fromMLinesL n bsToIntTriple

mLinesToIntV :: Int -> IO (VU.Vector Int)
mLinesToIntV n = fromMLinesV n bsToInt

mLinesToDoubleV :: Int -> IO (VU.Vector Double)
mLinesToDoubleV n = fromMLinesV n bsToDouble

mLinesToTupleV :: Int -> IO (VU.Vector (Int, Int))
mLinesToTupleV n = fromMLinesV n bsToIntTuple
    
mLinesToTripleV :: Int -> IO (VU.Vector (Int, Int, Int))
mLinesToTripleV n = fromMLinesV n bsToIntTriple

-- output
yesnoL :: Bool -> IO ()
yesnoL cond
    | cond = putStrLn "Yes"
    | otherwise = putStrLn "No"

yesnoU :: Bool -> IO ()
yesnoU cond
    | cond = putStrLn "YES"
    | otherwise = putStrLn "NO"

-- others
strToChr :: String -> Char
strToChr [x] = x

chrToStr :: Char -> String
chrToStr x = [x]

m1 :: Int -> Int
m1 x = subtract 1 x

toDouble :: Int -> Double
toDouble x = fromIntegral x

toInt :: Double -> Int
toInt x = floor x

sqToList :: SQ.Seq a -> [a]
sqToList sq = 
    case SQ.viewl sq of
        SQ.EmptyL -> []
        x SQ.:< sq' -> x : sqToList sq'