p03989 AtCoder Grand Contest 005 - ~K Perm Counting
Snuke loves permutations. He is making a permutation of length N.

Since he hates the integer K, his permutation will satisfy the following:

* Let the permutation be a_1, a_2, ..., a_N. For each i = 1,2,...,N, |a_i - i| \neq K.



Among the N! permutations of length N, how many satisfies this condition?

Since the answer may be extremely large, find the answer modulo 924844033(prime).

Constraints

* 2 ≦ N ≦ 2000
* 1 ≦ K ≦ N-1

Input

The input is given from Standard Input in the following format:


N K


Output

Print the answer modulo 924844033.

Examples

Input

3 1


Output

2


Input

4 1


Output

5


Input

4 2


Output

9


Input

4 3


Output

14


Input

425 48


Output

756765083
```python
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce


sys.setrecursionlimit(2147483647)
INF = 10 ** 13
def LI(): return list(map(int, sys.stdin.readline().split()))
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.readline().rstrip().split()
def S(): return sys.stdin.readline().rstrip()
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)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]

mod=924844033

n,k=LI()
fac = [1] * (n + 1)
for j in range(1, n + 1):
    fac[j] = fac[j-1] * j % mod

dp=[[0]*2 for _ in range(n+1)]
dp[0][0]=1
last=0
for i in range(min(n,2*k)):
    idx=i
    while idx<n:
        ndp = [[0] * 2 for _ in range(n+1)]
        if idx==i:
            for ll in range(n+1):
                dp[ll][0]+=dp[ll][1]
                dp[ll][1]=0
        for l in range(1,n+1):
            ndp[l][0]=sum(dp[l])
            if idx-k>=0:
                ndp[l][0]+=dp[l-1][0]
            if idx+k<n:
                ndp[l][1]=sum(dp[l-1])
            ndp[l][0]%=mod
        ndp[0][0]=1
        dp=ndp
        idx+=2*k

ans=fac[n]
for m in range(1,n+1):
    if m%2:
        ans-=sum(dp[m])*fac[n-m]%mod
    else:
        ans+=sum(dp[m])*fac[n-m]%mod
    ans%=mod

print(ans)











```