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 Lua (5.3.2) (judge verdict: Wrong Answer)

local function use_strict()
    local mt = {
        __index = function(t,k) error("_G __index:" .. k) end,
        __newindex = function(t,k,v) error("_G __newindex:" .. k) end
    }
    setmetatable(_G, mt)
end
local DBG = false
local function dbgpr(...)
    if DBG then
        io.write("[dbg]")
        print(...)
    end
end

local function dbgpr_t(tbl, use_pairs)
    if DBG then
        local enum = ipairs
        if use_pairs then
            enum = pairs
        end
        dbgpr(tbl)
        io.write("[dbg]")
        for i,v in enum(tbl) do
            io.write(i)
            io.write(":")
            io.write(tostring(v))
            io.write(" ")
        end
        print("")
    end
end

local function dbgpr_tp(tbl)
    dbgpr_t(tbl, true)
end

local function str2tbl(s)
    local t = {string.byte(s, 1, #s)}
    for i=1, #t do
        t[i] = string.char(t[i])
    end
    return t
end

--
use_strict()
local YES, NO = "YES", "NO"

local H, W, N = io.read("*n", "*n", "*n", "*l")
local sr, sc = io.read("*n", "*n", "*l")
local S = str2tbl(io.read("*l"))
local T = str2tbl(io.read("*l"))

local opposite = {
    L="R",
    R="L",
    D="U",
    U="D"
}

local function solve(size, move_lower, init_pos)
    local direc = move_lower == "L" and "horiz" or "verti"
    local move_upper = opposite[move_lower]
    -- Aoki wins when position p satisfies lower_bound <= p <= upper_bound
    local lower_bound = 1
    local upper_bound = size
    local math_max = math.max
    local math_min = math.min
    for n=N,1,-1 do
        if T[n] == move_lower then
            upper_bound = math_min(upper_bound + 1, size)
        elseif T[n] == move_upper then
            lower_bound = math_max(lower_bound - 1, 1)
        end
        if S[n] == move_lower then
            lower_bound = lower_bound + 1
        elseif S[n] == move_upper then
            upper_bound = upper_bound - 1
        end
        if lower_bound > size then
            return NO
        end
        if upper_bound < 1 then
            return NO
        end
        if lower_bound > upper_bound then
            return NO
        end
        dbgpr(direc, n, S[n], T[n], lower_bound, upper_bound)
    end
    if lower_bound <= init_pos and init_pos <= upper_bound then
        return YES
    else
        return NO
    end
end

local horiz_ans = solve(W, "L", sc)
dbgpr("horiz", horiz_ans)
if horiz_ans == NO then
    print(NO)
    return
end
local vert_ans = solve(H, "D", sr)
dbgpr("vert", vert_ans)
if vert_ans == NO then
    print(NO)
    return
end
print(YES)