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# include <bits/stdc++.h>
using namespace std ;
# define int long long
# define endl "\n"
const int MOD = 5000011 ; // 质数，可以使用费马小定理求逆元
const int N = 1e5 + 10 ;
int fact [ N ] ; // 阶乘数组
// 快速幂
int qmi ( int a , int b , int p ) {
int res = 1 ;
a % = p ;
while ( b ) {
if ( b & 1 ) res = res * a % p ;
b > > = 1 ;
a = a * a % p ;
}
return res ;
}
// 利用组合数定义式求C(m,n),用到快速幂求逆元
int C ( int m , int n ) {
if ( m < n ) return 0 ;
return fact [ m ] * qmi ( fact [ n ] , MOD - 2 , MOD ) % MOD * qmi ( fact [ m - n ] , MOD - 2 , MOD ) % MOD ;
}
int n , k ; // n只牛,两只牡牛之间，
signed main ( ) {
// 预处理范围内数字的阶乘
fact [ 0 ] = 1 ;
for ( int i = 1 ; i < N ; i + + ) fact [ i ] = ( fact [ i - 1 ] * i ) % MOD ;
cin > > n > > k ;
int res = 0 ;
for ( int i = 1 ; i < = ( n + k ) / ( k + 1 ) ; i + + ) // 枚举牡牛的所有可能数量
res = ( res + C ( n - ( i - 1 ) * k , i ) ) % MOD ; // 累计组合数
// 如果0只牡牛,还应该增加一种方案，就是全都是牝牛
cout < < res + 1 < < endl ;
}
