思路：

$\Pi_{i=1}^n\Pi_{j=1}^m f[gcd(i,j)]$

$=\Pi_{d=1}^n\Pi_{i=1}^{\lfloor\frac{n}{d}\rfloor}\Pi_{j=1}^{\lfloor\frac{m}{d}\rfloor}f[d]*(gcd(i,j)==1)$

$\Sigma_{k=1}^n\Pi_{d|k}\Pi_{i=1}^{\lfloor\frac{n}{dk}\rfloor}\Pi_{j=1}^{\lfloor\frac{m}{dk}\rfloor}*f[d]*\mu(k)$

设dk=t

$=\Sigma_{t=1}^n\Pi_{i=1}^{\lfloor\frac{n}{t}\rfloor}\Pi_{j=1}^{\lfloor\frac{m}{t}\rfloor}\Pi_{d|t}f[d]*\mu(\frac{t}{d})$

$=\Sigma_{t=1}^n\Pi_{d|t}f[d]^{\mu(\frac{t}{d})*\lfloor\frac{n}{t}\rfloor*\lfloor\frac{m}{t}\rfloor}$

设$g(t)=\Pi_{d|t}f[d]^{\mu(\frac{t}{d})}$

$g(t)可以O(nlogn)预处理$

搞个前缀积

剩下的 喜闻乐见 分块

//By SiriusRen#include 
     
      #include 
      
       #include 
       
        using namespace std;const int N=1000005,mod=1000000007;int n,m,cases,f[N],vis[N],prime[N],g[N],mu[N],tot;typedef long long ll;int pow(ll x,int y){    ll res=1;    while(y){        if(y&1)res=res*x%mod;        x=x*x%mod,y>>=1;    }return res;}void shai(){    mu[1]=f[1]=g[0]=1;    for(int i=2;i
        
         m)swap(n,m);        ll ans=1;        for(int l=1,r;l<=n;l=r+1){            r=min(n/(n/l),m/(m/l));            ans=ans*pow(1ll*g[r]*pow(g[l-1],mod-2)%mod,1ll*(n/l)*(m/l)%(mod-1))%mod;        }printf("%lld\n",ans);    }}