"""

Tea Time (5) 想定解

O( max(N,K) ) Ver

"""

import sys
from sys import stdin

mod = 998244353

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

fac,inv = modfac(1000000,mod)

N,K = map(int,stdin.readline().split())

assert 1 <= N <= 200000
assert 0 <= K <= 200000

if K == 0:
    print (1)
    sys.exit()

# 括弧列のpeakがK個ある場合を考える
Kinv = pow(K,mod-2,mod)
ans = 0

for peak in range(1,K+1):

    # 括弧列の個数
    bracket_num = modnCr(K,peak) * modnCr(K,peak-1) * Kinv % mod

    # 隣り合う相異なる記号の個数
    diff_pos = peak*2-1

    ans += modnCr(2*K+N-diff_pos,N-diff_pos) * bracket_num
    ans %= mod

print (ans % mod)