Problem:
Score : 600 points

Problem Statement

We have a rectangular grid of squares with H horizontal rows and W vertical columns. Let (i,j) denote the square at the i-th row from the top and the j-th column from the left.
On this grid, there is a piece, which is initially placed at square (s_r,s_c).

Takahashi and Aoki will play a game, where each player has a string of length N.
Takahashi's string is S, and Aoki's string is T. S and T both consist of four kinds of letters: L, R, U and D.

The game consists of N steps. The i-th step proceeds as follows:

First, Takahashi performs a move. He either moves the piece in the direction of S_i, or does not move the piece.

Second, Aoki performs a move. He either moves the piece in the direction of T_i, or does not move the piece.

Here, to move the piece in the direction of L, R, U and D, is to move the piece from square (r,c) to square (r,c-1), (r,c+1), (r-1,c) and (r+1,c), respectively. If the destination square does not exist, the piece is removed from the grid, and the game ends, even if less than N steps are done.

Takahashi wants to remove the piece from the grid in one of the N steps.
Aoki, on the other hand, wants to finish the N steps with the piece remaining on the grid.
Determine if the piece will remain on the grid at the end of the game when both players play optimally.

Constraints

2 \leq H,W \leq 2 \times 10^5

2 \leq N \leq 2 \times 10^5

1 \leq s_r \leq H

1 \leq s_c \leq W

|S|=|T|=N

S and T consists of the four kinds of letters L, R, U and D.

Input

Input is given from Standard Input in the following format:

H W N
s_r s_c
S
T

Output

If the piece will remain on the grid at the end of the game, print YES; otherwise, print NO.

Sample Input 1

2 3 3
2 2
RRL
LUD

Sample Output 1

YES

Here is one possible progress of the game:

Takahashi moves the piece right. The piece is now at (2,3).

Aoki moves the piece left. The piece is now at (2,2).

Takahashi does not move the piece. The piece remains at (2,2).

Aoki moves the piece up. The piece is now at (1,2).

Takahashi moves the piece left. The piece is now at (1,1).

Aoki does not move the piece. The piece remains at (1,1).

Sample Input 2

4 3 5
2 2
UDRRR
LLDUD

Sample Output 2

NO

Sample Input 3

5 6 11
2 1
RLDRRUDDLRL
URRDRLLDLRD

Sample Output 3

NO

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

{-# LANGUAGE
  ScopedTypeVariables, BangPatterns, TupleSections, ExplicitForAll,
  LambdaCase, MultiWayIf, Unsafe, RecordWildCards, FlexibleContexts, CPP,
  NoMonomorphismRestriction, GADTs, PatternGuards,
  RankNTypes, EmptyDataDecls, EmptyCase, ViewPatterns #-}


import Data.Bits
import Data.List
import Data.Maybe
import Data.Tuple
import Data.Ord
import Data.Int
import Data.Word
import Data.Char
import Data.Function
import Data.STRef
import Data.IORef
import Data.Monoid
import Data.Functor
import System.IO
import Control.Arrow
import Control.Applicative
import Control.Monad
import Control.Monad.Primitive
import Control.Monad.State.Strict
import Control.Monad.ST
import Control.Monad.ST.Lazy (strictToLazyST, lazyToStrictST)
import qualified Control.Monad.ST.Lazy as STL
-- import Control.Monad.ST.Safe
import Control.DeepSeq
import qualified Data.ByteString.Char8 as BS
import qualified Data.ByteString.Lazy.Char8 as BSL
import qualified Data.ByteString.Builder as BSB
import Data.IntMap (IntMap)
import qualified Data.IntMap.Strict as IMS
import qualified Data.IntMap.Lazy as IML
import Data.IntSet (IntSet)
import qualified Data.IntSet as IS
import qualified Data.Sequence as Seq
import Data.Sequence (Seq)
import qualified Data.Array.IArray as A
import qualified Data.Array.MArray.Safe as A
import qualified Data.Array.MArray as A
import Data.Array (Array)
import Data.Array.Unboxed (UArray)
import Data.Array.IArray (IArray)
import Data.Array.MArray.Safe (MArray)
import Data.Array.IO.Safe (IOArray, IOUArray)
import Data.Array.ST.Safe (STArray, STUArray, runSTArray, runSTUArray)
import qualified Data.Vector as V
import qualified Data.Vector.Mutable as VM
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import qualified Data.Vector.Primitive as VP
import qualified Data.Vector.Primitive.Mutable as VPM
import qualified Data.Vector.Generic as VG
import qualified Data.Vector.Generic.Mutable as VGM
import qualified Data.Attoparsec.ByteString.Char8 as Atto



main :: IO ()
main = do
  [!h,!w,!n] <- map readInt . words <$> getLine
  [!or,!oc] <- map readInt . words <$> getLine
  s <- VU.unfoldrN n BS.uncons <$> BS.getLine
  t <- VU.unfoldrN n BS.uncons <$> BS.getLine
  if VU.foldr (\(s,t) cont (!l0,!r0,!u0,!d0) ->
                 let (!l1,!r1,!u1,!d1) = case s of
                       'L' -> (l0-1,r0,u0,d0)
                       'R' -> (l0,r0+1,u0,d0)
                       'U' -> (l0,r0,u0-1,d0)
                       'D' -> (l0,r0,u0,d0+1)
                 in if l1 < 1 || r1 > w || u1 < 1 || d1 > h then True
                    else cont $! force
                         $ case t of
                             'R' | l1 < w    -> (l1+1,r1,u1,d1)
                                 | otherwise -> (l1,r1,u1,d1)
                             'L' | r1 > 1    -> (l1,r1-1,u1,d1)
                                 | otherwise -> (l1,r1,u1,d1)
                             'D' | u1 < h    -> (l1,r1,u1+1,d1)
                                 | otherwise -> (l1,r1,u1,d1)
                             'U' | d1 > 1    -> (l1,r1,u1,d1-1)
                                 | otherwise -> (l1,r1,u1,d1))
     (\(!l,!r,!u,!d) -> False) (VU.zip s t) (oc,oc,or,or)
    then putStrLn "NO" else putStrLn "YES"
                             
                                           
              

countLRUD :: BS.ByteString -> (Int,Int,Int,Int)
countLRUD =
  force . BS.foldl' (\(!l,!r,!u,!d) ch -> case ch of
                        'L' -> (l+1,r,u,d)
                        'R' -> (l,r+1,u,d)
                        'U' -> (l,r,u+1,d)
                        'D' -> (l,r,u,d+1)) (0,0,0,0)

#define IL(f) {-# INLINE f #-}; f

rInt :: StateT BSL.ByteString Maybe Int
IL(rInt) = StateT $ BSL.readInt . BSL.dropWhile (<'!')
rIntS :: StateT BS.ByteString Maybe Int
IL(rIntS) = StateT $ BS.readInt . BS.dropWhile (<'!')

-- unlessM, whenM :: (Monad m) => m Bool -> m () -> m ()
IL(whenM) = (. flip when) . (>>=)
IL(unlessM) = (. flip unless) . (>>=)

IL(wrA) = A.writeArray
IL(rdA) = A.readArray
IL(mdA) = \arr f !i -> do
  ai <- rdA arr i
  let fai = f ai 
  wrA arr i fai
  return (ai,fai)
{-# INLINE mdA' #-}
mdA' = \arr f !i -> do
  !ai <- rdA arr i
  let !fai = f ai
  wrA arr i fai
  return (ai,fai)
IL(swapA) = \arr !i !j -> do
  ai <- rdA arr i
  wrA arr i =<< rdA arr j
  wrA arr j ai

#define D(f,r,d)\
  IL(f) :: Integral a=>a->d; f=fromIntegral;\
  IL(r) :: String->d; r=read
#define C(f,r,g,h,d) D(f,r,d);\
  g,h :: RealFrac a=>a->d; IL(g)=floor; IL(h)=ceiling
C(_toInteger_,readInteger,floorInteger,ceilInteger,Integer)
C(toInt,readInt,floorInt,ceilInt,Int)
C(toI8,readI8,floorI8,ceilI8,Int8)
C(toI16,readI16,floorI16,ceilI16,Int16)
C(toI32,readI32,floorI32,ceilI32,Int32)
C(toI64,readI64,floorI64,ceilI64,Int64)
C(toWord,readWord,floorWord,ceilWord,Word)
C(toW8,readW8,floorW8,ceilW8,Word8)
C(toW16,readW16,floorW16,ceilW16,Word16)
C(toW32,readW32,floorW32,ceilW32,Word32)
C(toW64,readW64,floorW64,ceilW64,Word64)
D(toDouble,readDouble,Double)
D(toFloat,readFloat,Float)
#undef D
#undef C

#define N(f,g,h,a,m)\
  IL(f) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> e -> m (a i e);\
  f=A.newArray;\
  IL(g) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> m (a i e);\
  g=A.newArray_;\
  IL(h) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> [e] -> m (a i e);\
  h=A.newListArray
#define C(a,m)
N(newIOA,newIOA_,newIOAL,IOArray,IO)
N(newSTA,newSTA_,newSTAL,STArray s,ST s)
#undef C
#define C(a,m) MArray (a) e (m), 
N(newIOUA,newIOUA_,newIOUAL,IOUArray,IO)
N(newSTUA,newSTUA_,newSTUAL,STUArray s,ST s)
#undef C
#undef N

#undef IL