"""

Tea Time (5) 想定解

定数倍高速化Ver

注: O(N+K^3) なので、通りません

"""

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(1000,mod)


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

# 制約チェック
assert 1 <= N <= 400
assert 0 <= K <= 400

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

# 1つ目の括弧は置いた状態から開始
dp = [[[0],[0]] for l in range(2)]
dp[0][1][0] = 1 #last,s,p の順番

for i in range(2,2*K+1):

    ndp = [[[0] * i for nnum in range(min(K+1,i+1))] for l in range(2)]

    for put in range(2): #置く括弧 0 = ( , 1 = )
        for last in range(2):
            put_neq_last = (1 if last != put else 0)
            for ls in range(len(dp[0])):
                news = ls + (1,-1)[put]
                if 0 <= news < len(ndp[0]):
                    for lp in range(len(dp[0][0])):
                        newp = lp + put_neq_last
                        ndp[put][news][newp] = (ndp[put][news][newp] + dp[last][ls][lp]) % mod
    
    dp = ndp

    #print (dp)
ans = 0

for p in range(len(dp[0][0])):

    now = dp[1][0][p] * modnCr(2*K+N-p,N-p)
    #print (p,dp[0][p][1] , modnCr(2*K+N-p,N-p))
    ans += now
    ans %= mod

print (ans)