Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int mod = 924844033, N = 2e3 + 1;
#define For( i, j, k ) for( int i = j; i <= k; ++i )
int n, k, cnt, f[ N * 2 ][ N ][ 2 ], fac[ N ];
bool link[ N * 2 ];
int main( )
{
 std::cin >> n >> k;
 f[ 0 ][ 0 ][ 0 ] = 1;
 fac[ 0 ] = 1;
 For( i, 1, n )
  fac[ i ] = ( long long ) fac[ i - 1 ] * i % mod;
 for( int i = 1; i <= k; ++i ) for( int t = 1; t <= 2; ++t )
 for( int j = i; j <= n; j += k )
 {
  ++cnt;
  if( j != i )
   link[ cnt ] = 1;
 }
 for( int i = 1; i <= cnt; ++i )
 {
  for( int j = 0; j <= n; ++j )
  {
   f[ i ][ j ][ 0 ] = ( f[ i - 1 ][ j ][ 0 ] + f[ i - 1 ][ j ][ 1 ] ) % mod;
   if( link[ i ] && j )
    f[ i ][ j ][ 1 ] = f[ i - 1 ][ j - 1 ][ 0 ];
  }
 }
 int ans = 0;
 for( int i = 0; i <= n; ++i )
 {
  if( i & 1 )
   ans = ( ans - ( long long ) fac[ n - i ] * ( f[ n * 2 ][ i ][ 1 ] + f[ n * 2 ][ i ][ 0 ] ) % mod + mod ) % mod;
  else
   ans = ( ans + ( long long ) fac[ n - i ] * ( f[ n * 2 ][ i ][ 1 ] + f[ n * 2 ][ i ][ 0 ] ) % mod ) % mod;
 }
 std::cout << ans;
 return 0;
}