DZY Loves Math VI

求下面表达式的值，对 \(10^9 + 7\) 取模。

\[\sum\limits_{i=1}^{n} \sum\limits_{j=1}^{m} \operatorname*{lcm}(i,j)^{\gcd(i,j)} \]

\(1 \leq n,m \leq 5 \times 10^5\)。

---

\[\begin{aligned} & \sum\limits_{i=1}^{n} \sum\limits_{j=1}^{m} \operatorname*{lcm}(i,j)^{\gcd(i,j)} \\ = & \sum\limits_{d=1}^{\min(n,m)} \sum\limits_{i=1}^{\lfloor \frac{n}{d} \rfloor} \sum\limits_{j=1}^{\lfloor \frac{m}{d} \rfloor} (d \times i \times j)^d [\gcd(i,j)=1] \\ = & \sum\limits_{d=1}^{\min(n,m)} d^d \sum\limits_{i=1}^{\lfloor \frac{n}{d} \rfloor} \sum\limits_{j=1}^{\lfloor \frac{m}{d} \rfloor} i^d \times j^d \times \sum\limits_{e \mid \gcd(i,j)} \mu(e) \\ = & \sum\limits_{d=1}^{\min(n,m)} d^d \sum\limits_{e=1}^{\lfloor \frac{\min(n,m)}{d} \rfloor} \mu(e) \times e^{2d} \times \sum\limits_{i=1}^{\lfloor \frac{n}{de} \rfloor} \sum\limits_{j=1}^{\lfloor \frac{m}{de} \rfloor} i^d \times j^d \\ = & \sum\limits_{d=1}^{\min(n,m)} d^d \sum\limits_{e=1}^{\lfloor \frac{\min(n,m)}{d} \rfloor} \mu(e) \times e^{2d} \times (\sum\limits_{i=1}^{\lfloor \frac{n}{de} \rfloor} i^d) \times (\sum\limits_{i=1}^{\lfloor \frac{m}{de} \rfloor} i^d) \\ \end{aligned} \]

// Ad astra per aspera
#include<iostream>
#include<cstdio>
#include<vector>
using namespace std;
const long long mod=1000000007;
const int V=100000;
long long mu[500010],pre[500010];
long long pow_mod(long long num1,long long num2){
	long long num3=1;
	while(num2){
		if(num2&1){
			num3=num3*num1%mod;
		}
		num1=num1*num1%mod;
		num2>>=1;
	}
	return num3;
} 
int main(){
	int n,m;
	scanf("%d %d",&n,&m);
	int small=min(n,m),big=max(n,m);
	mu[1]=1;
	for(int i=1;i<=small;i++){
		for(int j=2;j<=small/i;j++){
			mu[i*j]-=mu[i];
		}
	}
	long long ans=0;
	for(int d=1;d<=small;d++){
		for(int i=1;i<=big/d;i++){
			pre[i]=(pre[i-1]+pow_mod(i,d))%mod;
		}
		long long sum=0;
		for(int e=1;e<=small/d;e++){
			sum+=mu[e]*pow_mod(e,2*d)%mod*pre[n/d/e]%mod*pre[m/d/e]%mod;
			sum%=mod;
		}
		ans+=sum*pow_mod(d,d)%mod;
		ans%=mod;
	}
	printf("%lld",ans);
    return 0;
}