hdu 2865 基礎polya問題

#include<cstdio>
#include<iostream>
using namespace std;
const int MOD = 1000000007;
typedef long long LL;
LL n, m;
int prime[37000],is[37000];
void getprime(){
	int cnt=0;
	for(int i=2;i<37000;i++){
		if(!is[i]){
			prime[cnt++]=i;
			for(int j=i;j<37000;j+=i)
				is[j]=1;
		}
	}
}
LL pow(LL x,LL n)
{
	LL t=1;
	while(n){
		if(n&1)t=(t*x)%MOD;
		x=(x*x)%MOD;
		n >>= 1;
	}
	return t%MOD;
}
LL getptr(LL p,LL i)
{
	LL tmp=pow(p-1,i);
	if(i&1)
		tmp=(tmp+MOD-(p-1))%MOD;
	else
		tmp=(tmp+p-1)%MOD;
	return tmp;
}
LL eular(LL x)
{
	if(x==1)return 1;
	LL rep=x;
	for(int i=0;prime[i]*prime[i]<=x;i++){
		if(x%prime[i]==0){
			rep-=rep/prime[i];
			x/=prime[i];
		}
		while(x%prime[i]==0)
			x/=prime[i];
		if(x==1)break;
	}
	if(x!=1)rep-=rep/x;
	return rep%MOD;
}
LL inv(LL n)
{
	return pow(n,MOD-2)%MOD;
}
int main(){
	getprime();
	while(cin>>n>>m){
		LL ans=0;
		int  i;
		m--;
		for(i = 1;i * i < n;i++){
			if(n%i==0){
				ans=(ans+(getptr(m,i)*eular(n/i))%MOD)%MOD;
				ans=(ans+(getptr(m,n/i)*eular(i))%MOD)%MOD;
			}
		}
		if(i*i==n)
			ans=(ans+(getptr(m,i)*eular(n/i))%MOD)%MOD;
		ans=((m+1)*ans)%MOD;
		ans=(ans*inv(n))%MOD;
		cout<<ans<<endl;
	}
}