逆元（清北 算）

 

#include<cstdio>
#define ll long long
int p=1e9+7;

ll fastpow(ll x,ll k)
{
    ll b=1;
    while (k>0)
    {
          if (k%2==1)
             b=(b*x)%p;
          k=k/2;
          x=(x*x)%p;
    }
    b=b%p;
    return b;
} 
int n,m;
int main(){
 
    scanf("%d%d",&n,&m);ll sum=m;
    
   
    
        for(int i=2;i<=n;i++)
    {
        ll w=fastpow(i,m);
        
        sum+=((i*((w-1+p)%p))%p*fastpow(i-1,p-2))%p;
        
        sum%=p;
	
    }
    printf("%d",int(sum));
    
    
}

ax ≡ 1 (mod b)

b^-1x≡1

a,b互质

x^b-1≡ 1(mod b)

a*x^-1(mod b)=a*x^b-2(mod b)