p03270 AtCoder Regular Contest 102 - Stop. Otherwise...
Takahashi throws N dice, each having K sides with all integers from 1 to K. The dice are NOT pairwise distinguishable. For each i=2,3,...,2K, find the following value modulo 998244353:

* The number of combinations of N sides shown by the dice such that the sum of no two different sides is i.



Note that the dice are NOT distinguishable, that is, two combinations are considered different when there exists an integer k such that the number of dice showing k is different in those two.

Constraints

* 1 \leq K \leq 2000
* 2 \leq N \leq 2000
* K and N are integers.

Input

Input is given from Standard Input in the following format:


K N


Output

Print 2K-1 integers. The t-th of them (1\leq t\leq 2K-1) should be the answer for i=t+1.

Examples

Input

3 3


Output

7
7
4
7
7


Input

4 5


Output

36
36
20
20
20
36
36


Input

6 1000


Output

149393349
149393349
668669001
668669001
4000002
4000002
4000002
668669001
668669001
149393349
149393349
```python
from sys import exit, setrecursionlimit, stderr
from functools import reduce
from itertools import *
from collections import defaultdict
from bisect import bisect

def read():
  return int(input())

def reads():
  return [int(x) for x in input().split()]

K, N = reads()
MOD = 998244353

MAXN = 4020

fact = [1] * MAXN
for i in range(1, MAXN): fact[i] = (fact[i-1] * i) % MOD
invfact = [0] * MAXN; invfact[-1] = pow(fact[-1], MOD-2, MOD)
for i in range(MAXN-2, -1, -1): invfact[i] = invfact[i+1] * (i+1) % MOD
def comb(n, k):
  return fact[n] * invfact[n-k] * invfact[k] % MOD if 0 <= k <= n else 0
def h(n, k):
  return comb(n+k-1, n)

def solve(n, k, u):
  ans = 0
  for q in range(u+1):
    ans = (ans + comb(u, q) * pow(2, u-q, MOD) % MOD * h(n-u+q, k-q-u)) % MOD
  return ans

anss = [0] * (2*K+1)
for i in range(2, K+2):
  u = sum(j < i-j <= K for j in range(1, K+1))
  if i % 2 == 1:
    ans = solve(N, K, u)
  else:
    ans = (solve(N, K-1, u) + solve(N-1, K-1, u)) % MOD
  anss[i] = ans
  print(ans)

for j in range(K+2, 2*K+1):
  print(anss[2*K - j + 2])

```