莫比乌斯反演

typedef long long ll;
const int MOD = 1e9+7;
#define mod(x) (ll(x)%MOD)
ll pow_m(ll a , ll n) {
	ll ret = 1;
	ll tmp = mod(a);
	while(n) {
		if(n&1) 
		ret = mod(ret*tmp);
		tmp = mod(tmp*tmp);
		n >>= 1;
	}
	return ret;
}

//计算1 + p + p^2 +...+p^n;

ll sum(ll p , ll n) {
	if(p == 0)
	return 0;
	if(n == 0)
	return 1;
	if(n & 1)
	return mod(mod(1 + pow_m(p , n/2+1 )) * mod(sum(p , n/2)));
	else
	return mod(mod(1+pow_m(p , n/2+1)) * mod(sum(p , n/2)) + mod(pow_m(p , n/2)));
}
//返回A^B的约数之和%MOD;
ll solve(ll A , ll B)　{
	getFactors(A);
	ll ans = 1;
	for(int i = 0; i < fatCnt; i++) {
		ans *= mod(sum(factor[1][0] , B*factor[i][1]));
		ans = mod(ans);
	}
	return ans;
}