"""

bitごとに解けばよい

a0,a1,b0,b1
個あった時

1のORで1に
0のANDで0になる

1の方が後に来る確率だけ求めればok
　

"""

import sys
from sys import stdin

mod = 998244353
def inv(x):
    return pow(x,mod-2,mod)

N,X,Y = map(int,stdin.readline().split())

A = list(map(int,stdin.readline().split()))
B = list(map(int,stdin.readline().split()))

AONE  = [0] * 18
BZERO = [0] * 18

for i in A:
    for j in range(18):
        if i % 2 == 1:
            AONE[j] += 1
        i //= 2

for i in B:
    for j in range(18):
        if i % 2 == 0:
            BZERO[j] += 1
        i //= 2

#print (AONE,BZERO)

ans = 0
N*=2

Xinv = inv(X)
Yinv = inv(Y)

for bit in range(18):

    a0,a1,b0,b1 = X-AONE[bit] , AONE[bit] , BZERO[bit] , Y - BZERO[bit]

    anot = a0 * Xinv
    bnot = b1 * Yinv
    
    now = 0

    able = 1
    for i in range(N-1,-1,-1):

        if i % 2 == 0: #PCT
            now += able * a1 * Xinv
            able *= anot
            now %= mod
            able %= mod
        else:
            able *= bnot
            able %= mod

    ans += now * (2**bit)

    #print (a0,a1,b0,b1,now)
    
    ans %= mod

ans *= pow(X*Y,N//2,mod)

print (ans % mod)