容易想到可以转化为一个有m堆石子，石子总数不超过n-m的阶梯博弈。阶梯博弈的结论是相当于只考虑奇数层石子的nim游戏。
nim和不为0不好算，于是用总方案数减掉nim和为0的方案数。然后考虑dp，按位考虑，设f[i][j]为已确定奇数石子堆的第i位及以上的放法后，保证当前异或和为0，剩下j个石子时的方案数。转移套一些组合数即可。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define P 1000000009
#define N 150100
char getc(){char c=getchar();while ((c<‘A‘||c>‘Z‘)&&(c<‘a‘||c>‘z‘)&&(c<‘0‘||c>‘9‘)) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
int x=0,f=1;char c=getchar();
while (c<‘0‘||c>‘9‘) {if (c==‘-‘) f=-1;c=getchar();}
while (c>=‘0‘&&c<=‘9‘) x=(x<<1)+(x<<3)+(c^48),c=getchar();
return x*f;
}
int n,m,fac[N],inv[N],f[20][N],ans=1;
int C(int n,int m){if (m>n) return 0;return 1ll*fac[n]*inv[m]%P*inv[n-m]%P;}
int F(int n,int m){return C(n+m-1,m-1);}
int main()
{
#ifndef ONLINE_JUDGE
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
const char LL[]="%I64d\n";
#else
const char LL[]="%lld\n";
#endif
n=read(),m=read();
fac[0]=1;for (int i=1;i<=n+m;i++) fac[i]=1ll*fac[i-1]*i%P;
inv[0]=inv[1]=1;for (int i=2;i<=n+m;i++) inv[i]=P-1ll*(P/i)*inv[P%i]%P;
for (int i=2;i<=n+m;i++) inv[i]=1ll*inv[i-1]*inv[i]%P;
ans=C(n,m);n-=m;
f[19][n]=1;
for (int i=18;~i;i--)
for (int j=0;j<=n;j++)
for (int k=0;j+(2*k<<i)<=n&&2*k<=(m+1>>1);k++)
f[i][j]=(f[i][j]+1ll*f[i+1][j+(2*k<<i)]*C(m+1>>1,2*k)%P)%P;
for (int i=0;i<=n;i++)
ans=(ans+P-1ll*f[0][i]*F(i,(m>>1)+1)%P)%P;
cout<<ans;
return 0;
}