Pushing the persimmon, the answer is this East (I am not familiar with Mobius inversion, indeed QAQ)

∑d=1ndd∑x=1⌊nd⌋μ(x)⋅x2d∑i=1⌊ndx⌋id∑j=1⌊mdx⌋jd\sum_{d=1}^n d^d \sum_{x=1}^{\lfloor \frac{n}{d} \rfloor} \mu(x) \cdot x^{2d} \sum_{i=1}^{\lfloor \frac{n}{dx} \rfloor} i^d \sum_{j=1}^{\lfloor \frac{m}{dx} \rfloor} j^dd=1∑n​ddx=1∑⌊dn​⌋​μ(x)⋅x2di=1∑⌊dxn​⌋​idj=1∑⌊dxm​⌋​jd

(So tired, this persimmon is so complicated that I don't write any more.)
Then the violence asks you to take a little care of the two things behind it, O(nlogn)O(nlogn)O(nlogn).

After the initial pre-processing, there are two more log s for the direct and fast power of the next two items. As a belief player, the result is handed in directly T and QAQ

//#pragma GCC optimize(2)
//#pragma GCC optimize(3)
#include<bits/stdc++.h>
#include<tr1/unordered_map>
#define LL long long
#define re register
#define fr(i,x,y) for(re int i=(x),z=y;i<=z;i++)
#define rf(i,x,y) for(int i=(x);i>=(y);i--)
#define frl(i,x,y) for(int i=(x);i<(y);i++)
using namespace std;
using namespace tr1;
const int N=500005;
const int INF=2147483647;
const int p=1e9+7;
int n,m;
int mu[N],b[N],L,pri[N];

void read(int &x){
	char ch=getchar();x=0;
	for(;ch<'0'||ch>'9';ch=getchar());
	for(;ch>='0'&&ch<='9';ch=getchar()) x=(x<<3)+(x<<1)+ch-'0';
}

inline void Add(LL &x,LL y){
	x+=y;
	while(x<0) x+=p;
	while(x>=p) x-=p;
}

inline LL qpow(LL a,int n){
	LL ans=1;
	for(LL sum=a;n;n>>=1,sum=sum*sum%p) if (n&1) ans=ans*sum%p;
	return ans;
}

void init(){
	mu[1]=1;
	frl(i,2,N){
		if (!b[i]) pri[++L]=i,mu[i]=-1;
		for(int j=1;j<=L&&i*pri[j]<N;j++){
			b[i*pri[j]]=1;
			if (i%pri[j]==0){
				mu[i*pri[j]]=0;
				break;
			}
			mu[i*pri[j]]=-mu[i];
		}
	}
}

LL sum[N],pw[N];
int main(){
	init();
	read(n);read(m);
	if (n>m) swap(n,m);
	LL ans=0;
	fr(i,1,m) pw[i]=1;
	fr(i,1,n){
		//LL sum1=0,sum2=0;
		fr(j,1,m/i) sum[j]=(sum[j-1]+pw[j]*j%p)%p,pw[j]=pw[j]*j%p;
		//Add(ans,sum1*sum2%p*qpow(i,i)%p);
		LL s=0;
		fr(j,1,n/i)
		 Add(s,mu[j]*qpow(sum[j]-sum[j-1],2)%p*sum[n/i/j]%p*sum[m/i/j]%p);
		Add(ans,s*qpow(i,i)%p);
	}
	cout<<ans<<endl;
	return 0;
}