先要推式子
[egin{aligned}
varphi(x)varphi(y)&=(xprod_{p|x}frac{p-1}p)(yprod_{p|y}frac{p-1}p)\
&=(prod_{p|xy}frac{p-1}p)(prod_{p|gcd(x,y)}frac{p-1}p)xy\
&=varphi(xy)varphi(gcd(x,y))xy
end{aligned}
\Rightarrow
varphi(xy)=frac{varphi(x)varphi(y)gcd(x,y)}{varphi(gcd(x,y))}\
]
接着堆式子
[egin{aligned}
ans&=sum_isum_{i}varphi(a_ia_j)dis(i,j)\
&=sum_{d=1}^nfrac{d}{varphi(d)}sum_isum_jvarphi(a_i)varphi(a_j)dis(i,j)[gcd(a_i,a_j)=d]
\
f(d)&=sum_isum_j...[gcd(a_i,a_j)=d]
\
g(d)&=sum_{d|x}f(x)\
&=sum_isum_j...[d|gcd(a_i,a_j)]\
&=sum_{d|a_i}sum_{d|a_j}varphi(a_i)varphi(a_j)(dep_i+dep_j-2dep_{lca(i,j)})\
&=sum_{d|a_i}varphi(a_i)sum_{d|a_j}varphi(a_j)(dep_i+dep_j-2dep_{lca(i,j)})\
&=2sum_{d|a_i}varphi(a_i)dep_isum_{d|a_j}varphi(a_j)-2sum_{d|a_i}varphi(a_i)sum_{d|a_j}varphi(a_j)*dep_{lca(i,j)}
\
f(d)&=sum_{d|x}mu(frac{x}d)g(x)=sum_{i=1}^{lfloorfrac{n}d
floor}mu(i)g(di)
\
ans&=sum_{d=1}^nfrac{d}{varphi(d)}f(d)\
&=sum_{d=1}^nfrac{d}{varphi(d)}sum_{i=1}^{lfloorfrac{n}d
floor}mu(i)g(di)\
&=sum_{T=1}^ng(T)sum_{d|T}mu(frac{T}d)frac{d}{varphi(d)}
\
h(T)&=sum_{d|T}mu(frac{T}d)frac{d}{varphi(d)}
\
end{aligned}
]
留意到(h(T))可以直接筛。而(g(T))也是可以“筛”的（建立虚树，自底向上的带权根径覆盖）,然后把这俩玩意儿数乘……
简直绝了，复杂度是(O(blog n))级别的
代码留坑……