import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

import numpy as np

MOD = 998244353

N,M = map(int,readline().split())
A = np.array(list(readline().rstrip())) - 48
B = np.array(list(readline().rstrip())) - 48
C = np.array(list(readline().rstrip())) - 48
D = np.array(list(readline().rstrip())) - 48

A,B,C,D

def cumprod(arr,MOD):
    L = len(arr); Lsq = int(L**.5+1)
    arr = np.resize(arr,Lsq**2).reshape(Lsq,Lsq)
    for n in range(1,Lsq):
        arr[:,n] *= arr[:,n-1]; arr[:,n] %= MOD
    for n in range(1,Lsq):
        arr[n] *= arr[n-1,-1]; arr[n] %= MOD
    return arr.ravel()[:L]

def make_fact(U,MOD):
    x = np.arange(U,dtype=np.int64); x[0] = 1
    fact = cumprod(x,MOD)
    x = np.arange(U,0,-1,dtype=np.int64); x[0] = pow(int(fact[-1]),MOD-2,MOD)
    fact_inv = cumprod(x,MOD)[::-1]
    return fact,fact_inv

U = 10**6
fact, fact_inv = make_fact(U,MOD)

def cnt_vertical(N,M,A,B,C,D):
    """
    ・ABが長さN、CDが長さM。
    ・初手に縦線を入れる場合。
    ・縦線の右端の位置に応じて場合分けして集計
    ・右端の位置を決めると、（左端の位置ごとの左側の入れ方の累積和） x （右側の入れ方の累積和）、適当に重みをつける必要あり
    """
    AB = A+B
    # ある場所を初めての縦線としたときに、左側にできる部分。上にある点の個数と下にある点の個数
    L = C.sum(); LU = A.cumsum() - A; LD = B.cumsum() - B
    if L >= 1:
        NL = fact[LU+LD+L-1] * fact_inv[LU+LD] % MOD * fact_inv[L-1] % MOD
    else:
        NL = np.zeros(N,dtype=np.int64)
        i = np.where(AB>0)[0][0]
        NL[i] = 1
        
    R = D.sum(); RU = A[::-1].cumsum() - A[::-1]; RD = B[::-1].cumsum() - B[::-1]
    NR = fact[RU+RD+R-1] * fact_inv[RU+RD] % MOD * fact_inv[R-1] % MOD; NR = NR[::-1]
    # とりあえず
    x = np.full(N,1,np.int64); x[AB==2] = 2; coef = cumprod(x,MOD)
    x = np.full(N,1,np.int64); x[AB==2] = (MOD+1)//2; coef_inv = cumprod(x,MOD)
    NL *= coef_inv; NR *= coef; NL %= MOD; NR %= MOD
    NL[AB==2] *= 2; NL[NL>=MOD] -= MOD
    NL[AB==0] = 0; NR[AB==0] = 0
    # sum(l<=r) NL[l] NR[r] を求めればよい。
    NL_cum = NL.cumsum() % MOD
    return (NL_cum * NR % MOD).sum() % MOD

def AB_only(N,A,B):
    x = np.ones(N,dtype=np.int64)
    x[A+B==2] = 2
    return cumprod(x,MOD)[-1]

def ABC_only(N,M,A,B,C):
    raise Exception
    x = cnt_vertical(N,M,A,B,np.zeros_like(C),C)
    x += cnt_vertical(M,N,C,np.zeros_like(C),A,B)
    return x%MOD

def F(N,M,A,B,C,D):
    # 問題の答えを出します。
    NA,NB,NC,ND = [np.count_nonzero(x) for x in [A,B,C,D]]
    if all(x != 0 for x in [NA,NB,NC,ND]):
        return (cnt_vertical(N,M,A,B,C,D) + cnt_vertical(M,N,C,D,A,B)) % MOD
    # A,Cに0列を寄せる
    if NA != 0:
        NA,NB = NB,NA; A,B = B,A
    if NC != 0:
        NC,ND = ND,NC; C,D = D,C
    if NB == 0:
        return AB_only(M,C,D)
    if ND == 0:
        return AB_only(N,A,B)
    # B,Dは0ではない
    if NA == 0 and NC == 0:
        # 2面B,Dのみ -> binom(B+D,B)
        return fact[NB+ND] * fact_inv[NB] % MOD * fact_inv[ND] % MOD
    if NA == 0:
        return ABC_only(M,N,C,D,B)
    if NC == 0:
        return ABC_only(N,M,A,B,D)

answer = F(N,M,A,B,C,D)
print(answer)

### FIXED CODE ###

import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines

import numpy as np

MOD = 998244353

N,M = map(int,readline().split())
A = np.array(list(readline().rstrip()),dtype=np.int64) - 48
B = np.array(list(readline().rstrip()),dtype=np.int64) - 48
C = np.array(list(readline().rstrip()),dtype=np.int64) - 48
D = np.array(list(readline().rstrip()),dtype=np.int64) - 48

def cumprod(arr,MOD):
    L = len(arr); Lsq = int(L**.5+1)
    arr = np.resize(arr,Lsq**2).reshape(Lsq,Lsq)
    for n in range(1,Lsq):
        arr[:,n] *= arr[:,n-1]; arr[:,n] %= MOD
    for n in range(1,Lsq):
        arr[n] *= arr[n-1,-1]; arr[n] %= MOD
    return arr.ravel()[:L]

def make_fact(U,MOD):
    x = np.arange(U,dtype=np.int64); x[0] = 1
    fact = cumprod(x,MOD)
    x = np.arange(U,0,-1,dtype=np.int64); x[0] = pow(int(fact[-1]),MOD-2,MOD)
    fact_inv = cumprod(x,MOD)[::-1]
    return fact,fact_inv

U = 10**6
fact, fact_inv = make_fact(U,MOD)

def cnt_vertical(N,M,A,B,C,D):
    """
    ・ABが長さN、CDが長さM。
    ・初手に縦線を入れる場合。
    ・縦線の右端の位置に応じて場合分けして集計
    ・右端の位置を決めると、（左端の位置ごとの左側の入れ方の累積和） x （右側の入れ方の累積和）、適当に重みをつける必要あり
    """
    AB = A+B
    # ある場所を初めての縦線としたときに、左側にできる部分。上にある点の個数と下にある点の個数
    L = C.sum(); LU = A.cumsum() - A; LD = B.cumsum() - B
    if L >= 1:
        NL = fact[LU+LD+L-1] * fact_inv[LU+LD] % MOD * fact_inv[L-1] % MOD
    else:
        NL = np.zeros(N,dtype=np.int64)
        i = np.where(AB>0)[0][0]
        NL[i] = 1
        
    R = D.sum(); RU = A[::-1].cumsum() - A[::-1]; RD = B[::-1].cumsum() - B[::-1]
    NR = fact[RU+RD+R-1] * fact_inv[RU+RD] % MOD * fact_inv[R-1] % MOD; NR = NR[::-1]

    x = np.full(N,1,np.int64); x[AB==2] = 2; coef = cumprod(x,MOD)
    x = np.full(N,1,np.int64); x[AB==2] = (MOD+1)//2; coef_inv = cumprod(x,MOD)
    NL *= coef_inv; NR *= coef; NL %= MOD; NR %= MOD
    NL[AB==2] *= 2; NL[NL>=MOD] -= MOD
    NL[AB==0] = 0; NR[AB==0] = 0
    # sum(l<=r) NL[l] NR[r] を求めればよい。
    NL_cum = NL.cumsum() % MOD
    return (NL_cum * NR % MOD).sum() % MOD

def AB_only(N,A,B):
    x = np.ones(N,dtype=np.int64)
    x[A+B==2] = 2
    return cumprod(x,MOD)[-1]

def ABC_only(N,M,A,B,C):
    AB = A+B
    R = C.sum(); RU = A[::-1].cumsum() - A[::-1]; RD = B[::-1].cumsum() - B[::-1]
    NR = fact[RU+RD+R-1] * fact_inv[RU+RD] % MOD * fact_inv[R-1] % MOD; NR = NR[::-1]
    NR[AB==0] = 0
    x = np.full(N,1,np.int64); x[AB==2] = 2; coef = cumprod(x,MOD)
    x=(coef * NR % MOD).sum()%MOD
    assert x == cnt_vertical(N,M,A,B,np.zeros_like(C),C)
    n = A.sum()+B.sum()+C.sum()-1; m = C.sum()-1
    y = fact[n] * fact_inv[m] % MOD * fact_inv[n-m] % MOD
    return (x+y)%MOD

def F(N,M,A,B,C,D):
    # 問題の答えを出します。
    NA,NB,NC,ND = [np.count_nonzero(x) for x in [A,B,C,D]]
    if all(x != 0 for x in [NA,NB,NC,ND]):
        return (cnt_vertical(N,M,A,B,C,D) + cnt_vertical(M,N,C,D,A,B)) % MOD
    # A,Cに0列を寄せる
    if NA != 0:
        NA,NB = NB,NA; A,B = B,A; C = C[::-1]; D = D[::-1]
    if NC != 0:
        NC,ND = ND,NC; C,D = D,C; A = A[::-1]; B = B[::-1]
    if NB == 0:
        return AB_only(M,C,D)
    if ND == 0:
        return AB_only(N,A,B)
    # B,Dは0ではない
    if NA == 0 and NC == 0:
        # 2面B,Dのみ -> binom(B+D,B)
        return fact[NB+ND] * fact_inv[NB] % MOD * fact_inv[ND] % MOD
    if NA == 0:
        return ABC_only(M,N,C,D,B)
    if NC == 0:
        return ABC_only(N,M,A,B,D)

answer = F(N,M,A,B,C,D)
print(answer)

N=3
M=2
while True:
    A=np.random.randint(0,2,N).astype(np.int64)
    B=np.random.randint(0,2,N).astype(np.int64)
    C=np.random.randint(0,2,M).astype(np.int64)
    D=np.random.randint(0,2,M).astype(np.int64)
    x=F(N,M,A,B,C,D)
    y=F(M,N,C,D,A,B)
    z=F(N,M,B,A,C,D)
    if x != z:
        break

A,B,C,D

x,y,z

cnt_vertical(N,M,A,B,C,D),cnt_vertical(N,M,B,A,C,D)

cnt_vertical(M,N,C,D,A,B),cnt_vertical(M,N,C,D,B,A)