Problem:
Score : 200 points

Problem Statement

There are N participants in the CODE FESTIVAL 2016 Qualification contests. The participants are either students in Japan, students from overseas, or neither of these.

Only Japanese students or overseas students can pass the Qualification contests. The students pass when they satisfy the conditions listed below, from the top rank down. Participants who are not students cannot pass the Qualification contests.

A Japanese student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B.

An overseas student passes the Qualification contests if the number of the participants who have already definitively passed is currently fewer than A+B and the student ranks B-th or above among all overseas students.

A string S is assigned indicating attributes of all participants. If the i-th character of string S is a, this means the participant ranked i-th in the Qualification contests is a Japanese student; b means the participant ranked i-th is an overseas student; and c means the participant ranked i-th is neither of these.

Write a program that outputs for all the participants in descending rank either Yes if they passed the Qualification contests or No if they did not pass.

Constraints

1≦N,A,B≦100000

A+B≦N

S is N characters long.

S consists only of the letters a, b and c.

Input

Inputs are provided from Standard Input in the following form.

N A B
S

Output

Output N lines. On the i-th line, output Yes if the i-th participant passed the Qualification contests or No if that participant did not pass.

Sample Input 1

10 2 3
abccabaabb

Sample Output 1

Yes
Yes
No
No
Yes
Yes
Yes
No
No
No

The first, second, fifth, sixth, and seventh participants pass the Qualification contests.

Sample Input 2

12 5 2
cabbabaacaba

Sample Output 2

No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
No
No

The sixth participant is third among overseas students and thus does not pass the Qualification contests.

Sample Input 3

5 2 2
ccccc

Sample Output 3

No
No
No
No
No

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

--{-# LANGUAGE BangPatterns #-}
--{-# LANGUAGE ScopedTypeVariables #-}
--{-# LANGUAGE FlexibleContexts #-}
--{-# LANGUAGE ViewPatterns #-}
--{-# LANGUAGE OverloadedStrings #-}

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.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

--------------------------------------------------------------------------
-- 国内 : A+B
-- 海外 : A+B未満でかつB以内
solve :: String -> Int -> Int -> Int -> Int -> IO ()
solve [] _ _ _ _ = return ()
solve (x:xs) a b i j = 
    case x of
        'a' -> 
            if i < (a+b)
                then putStrLn "Yes" >> solve xs a b (i+1) j
                else putStrLn "No" >> solve xs a b i j
        'b' -> 
            if (i < (a+b)) && j < b
                then putStrLn "Yes" >> solve xs a b (i+1) (j+1)
                else putStrLn "No" >> solve xs a b i j
        'c' -> 
            putStrLn "No" >> solve xs a b i j

main = do
    [n,a,b] <- sLineToIntL
    xs <- str
    solve xs a b 0 0
--------------------------------------------------------------------------

-- Input Util
int :: IO Int
int = readLn 

double :: IO Double
double = readLn 

str :: IO String
str = getLine

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

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

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

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

sLineToIntL :: IO [Int]
sLineToIntL = strBS >>= return . map bsToInt . BC.words

sLineToDoubleL :: IO [Double]
sLineToDoubleL = strBS >>= return . map bsToDouble . BC.words

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

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

mLinesToIntL :: Int -> IO [Int]
mLinesToIntL n = replicateM n (strBS >>= return . bsToInt)

mLinesToDoubleL :: Int -> IO [Double]
mLinesToDoubleL n = replicateM n (strBS >>= return . bsToDouble)

mLinesToIntV :: Int -> IO (VU.Vector Int)
mLinesToIntV n = VU.replicateM n (strBS >>= return . bsToInt)

mLinesToDoubleV :: Int -> IO (VU.Vector Double)
mLinesToDoubleV n = VU.replicateM n (strBS >>= return . bsToDouble)

mLinesToTupleL :: Int -> IO [(Int, Int)]
mLinesToTupleL n = replicateM n (strBS >>= return . (\[a,b] -> (a,b)) . map bsToInt . BC.words)

mLinesToTripleL :: Int -> IO [(Int, Int, Int)]
mLinesToTripleL n = replicateM n (strBS >>= return . (\[a,b,c] -> (a,b,c)) . map bsToInt . BC.words)

mLinesToTupleV :: Int -> IO (VU.Vector (Int, Int))
mLinesToTupleV n = VU.replicateM n $ do
    bs <- strBS
    let
        Just (a, bs') = parseInt bs
        Just (b, _) = parseInt bs'
    return (a,b)
    
mLinesToTripleV :: Int -> IO (VU.Vector (Int, Int, Int))
mLinesToTripleV n = VU.replicateM n $ do
    bs <- strBS
    let
        Just (a, bs') = parseInt bs
        Just (b, bs'') = parseInt bs'
        Just (c, _) = parseInt bs''
    return (a,b,c)

-- 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