題意:
分析:
ll n,m;
ll hoppy,lad;
ll equip[maxn],frown[maxn];
ll qpow(ll hoppy1,ll lad1){
ll ans=1;
while(lad1){
if(lad1&1) ans=ans*hoppy1%mod;
lad1>>=1;
hoppy1=hoppy1*hoppy1%mod;
}
return ans;
}
void search(){
equip[0]=1;
frown[0]=1;
for(int i=1;i<=1e5;i++){
equip[i]=equip[i-1]*i%mod;
frown[i]=frown[i-1]*qpow(i,mod-2)%mod;
}
}
ll C(ll hoppy,ll lad){
return equip[hoppy]*frown[lad]%mod*frown[hoppy-lad]%mod;
}
void solve(){
if(n==0&&m==0) {
puts("1");
return ;
}
ll ans=0;
ll sum=(n+m)*2;
ans=C(sum,n+m);
if(m!=0)
ans=(ans-C(sum,m-1)+mod)%mod;
if(n!=0)
ans=(ans-C(sum,n-1)+mod)%mod;
cout<<ans<<endl;
}
int main(){
search();
while(~scanf("%lld%lld",&n,&m)) solve();
return 0;
}