We have a sequence of N integers: x=(x_0,x_1,\cdots,x_{N-1}). Initially, x_i=0 for each i (0 \leq i \leq N-1).

Snuke will perform the following operation exactly M times:

* Choose two distinct indices i, j (0 \leq i,j \leq N-1,\ i \neq j). Then, replace x_i with x_i+2 and x_j with x_j+1.



Find the number of different sequences that can result after M operations. Since it can be enormous, compute the count modulo 998244353.

Constraints

* 2 \leq N \leq 10^6
* 1 \leq M \leq 5 \times 10^5
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N M


Output

Print the number of different sequences that can result after M operations, modulo 998244353.

Examples

Input

2 2


Output

3


Input

3 2


Output

19


Input

10 10


Output

211428932


Input

100000 50000


Output

3463133
n,m = map(int,input().split())
mod = 998244353

#################################

## nCkのmodを求める関数
# テーブルを作る(前処理)
max = n+2*m + 100
fac, finv, inv = [0]*max, [0]*max, [0]*max

def comInit(max):
    fac[0] = fac[1] = 1
    finv[0] = finv[1] = 1
    inv[1] = 1

    for i in range(2,max):
      fac[i] = fac[i-1]* i% mod
      inv[i] = mod - inv[mod%i] * (mod // i) % mod
      finv[i] = finv[i-1] * inv[i] % mod

comInit(max)

# 二項係数の計算
def com(n,k):
    if(n < k):
        return 0
    if( (n<0) | (k < 0)):
        return 0
    return fac[n] * (finv[k] * finv[n-k] % mod) % mod


a = 0
for x in range(min(m,n)+1):
    if(3*m-x)%2==1:
        continue
    y = (3*m-x)//2
    a += com(n,x)*fac[y+n-1]*finv[y]*finv[n-1]
    a %= mod

b = fac[n-1+m-1] * finv[n-1] * finv[m-1] * n
b %= mod

ans = a-b
ans %= mod

print(ans)