"""

Merculialist (1)
想定解 解説準拠ver

"""

import sys
from sys import stdin

def modfac(n, MOD):
 
    f = 1
    factorials = [1]
    for m in range(1, n + 1):
        f *= m
        f %= MOD
        factorials.append(f)
    inv = pow(f, MOD - 2, MOD)
    invs = [1] * (n + 1)
    invs[n] = inv
    for m in range(n, 1, -1):
        inv *= m
        inv %= MOD
        invs[m - 1] = inv
    return factorials, invs

def modnCr(n,r):
    if n < 0 or r < 0 or n < r:
        return 0
    return fac[n] * inv[n-r] * inv[r] % mod

def inverse(x):
    return pow(x,mod-2,mod)

mod = 998244353
fac,inv = modfac(400000,mod)


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

assert 1 <= X <= 10**5
assert 1 <= Y <= 10**5
assert 1 <= Z <= 10**5
assert 1 <= K <= 10**5

N = X+Y+Z

# 全ての場合の数
all_case = (modnCr(N,X) * modnCr(N-X,Z) * fac[Y]) % mod


# 生きている場合の数を数えあげる
live_case = 0

for i in range(1,N+1): # i日目に最初にエリクシールを飲む場合

    # エリクシールの置き方
    xput = modnCr(N-i,X-1)

    yput = 1
    
    if i-K >= 1: #解説参照
        yput *= pow(N-i-X+K+1 , min(i-K,Y) , mod)

    if i-K < Y:
        L = N-X+1 - Y
        R = N-X+1 - max(0,i-K) - 1

        yput *= fac[R] * inv[L-1]
        yput %= mod
        
    live_case += xput * yput
    live_case %= mod

print (live_case * inverse(all_case) % mod)