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
#!usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
import sys
import math
import bisect
import random
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S(): return list(sys.stdin.readline())[:-1]
def IR(n):
    return [I() for i in range(n)]
def LIR(n):
    return [LI() for i in range(n)]
def SR(n):
    return [S() for i in range(n)]
def LSR(n):
    return [LS() for i in range(n)]

sys.setrecursionlimit(1000000)

#A
def A():
    n = I()
    if n <= 1000000000:
        print(0,0,1,0,0,n)
    else:
        b = 1000000000
        a = 1
        y = -n%1000000000
        x = (n+a*y)//b
        print(0,0,x,y,a,b)
        return

    return

#B
def B():
    n,k = LI()
    a = LI()
    s = []
    f = [1]*1000000
    for i in range(k):
        for j in range(n):
            aj = a[j]
            if f[aj]:
                s.append(aj)
                f[aj] = 0
            else:
                while 1:
                    x = s.pop(-1)
                    f[x] = 1
                    if x == aj:
                        break
        print(i+1,s)
    return

#C
mod = 998244353
def C():
    def comb(a,b):
        if a < b:
            return 1
        return fact[a]*inv[b]*inv[a-b]%mod
    n,m = LI()
    MA = 3*m+max(n,2*m)-1
    fact = [1]*(MA+1)
    for i in range(MA):
        fact[i+1] = fact[i]*(i+1)%mod
    inv = [1]*(MA+1)
    inv[MA] = pow(fact[MA],mod-2,mod)
    for i in range(MA)[::-1]:
        inv[i] = inv[i+1]*(i+1)%mod
    ans = 0
    for k in range(min(m,n)+1):
        if (3*m-k)%2 == 0:
            ans += comb(n,k)*comb((3*m-k)//2+n-1,n-1) #奇数の数がk個かつ総和が3mとなるパターンの総数(n個のうちk個に1を割り振る→総和が3m-kとなるパターンの総数を求める)
            ans %= mod
    ans -= n*comb(3*m-(2*m+1)+n-1,n-1)
    ans %= mod
    print(ans)

    return

#D
def D():
    return

#E
def E():
    n = I()

    return

#F
def F():
    n = I()

    return

#Solve
if __name__ == "__main__":
    C()
