SDOI2017数字表格

求$prod_{i=1}^nprod_{j=1}^n ext{Fib}[gcd(i,j)]; ext{mod};10^9+7$的值
令$nleq m$，则有：

egin{aligned}
prod_{i=1}^nprod_{j=1}^nf[gcd(i,j)]
&=prod_{d=1}^nprod_{i=1}^frac ndprod_{j=1}^frac md ext{Fib}[d]^{[gcd(i,j)=1]}\
&=prod_{d=1}^n ext{Fib}[d]^{sum_{i=1}^nsum_{j=1}^m[gcd(i,j)=d]}\
&=prod_{d=1}^n ext{Fib}[d]^{sum_{i=1}^{leftlfloorfrac nk ight floor}sum_{j=1}^{leftlfloorfrac mk ight floor}sum_{k|gcd(i,j)}mu(k)}\
&=prod_{d=1}^n ext{Fib}[d]^{sum_{i=1}^{leftlfloorfrac nk ight floor}sum_{j=1}^{leftlfloorfrac mk ight floor}sum_{k|i}sum_{k|j}mu(k)}\
&=prod_{d=1}^n ext{Fib}[d]^{sum_{i=1}^{leftlfloorfrac nk ight floor}sum_{k|i}sum_{j=1}^{leftlfloorfrac mk ight floor}sum_{k|j}mu(k)}\
&=prod_{d=1}^n ext{Fib}[d]^{sum_{i=1}^{minleft(leftlfloorfrac nk ight floor,leftlfloorfrac mk ight floor ight)}mu(k)sum_{i=1}^{leftlfloorfrac nk ight floor}sum_{k|i}sum_{j=1}^{leftlfloorfrac mk ight floor}sum_{k|j}1}\
&=prod_{d=1}^n ext{Fib}[d]^{sum_{i=1}^{minleft(leftlfloorfrac nk ight floor,leftlfloorfrac mk ight floor ight)}mu(k)sum_{i=1}^{leftlfloorfrac nk ight floor}sum_{k|i}1sum_{j=1}^{leftlfloorfrac mk ight floor}sum_{k|j}1}\
end{aligned}

...To be continue.