简化题意：给出一棵(n)个点的树，编号为(1)到(n)，第(i)个点的点权为(a_i)，保证序列(a_i)是一个(1)到(n)的排列，求
其中(dist(i,j))为树上(i,j)两点的距离。
看到(varphi)第一反应推式子
因为序列(a_i)是一个(1)到(n)的排列，设(t_i)表示点权为(i)的点的编号，那么原式等于$ frac{1}{n(n-1)} sumlimits_{i=1}^n sumlimits_{j=1}^n varphi(ij) dist(t_i,t_j)$
接下来化简$sumlimits_{i=1}^n sumlimits_{j=1}^nvarphi(ij) $
考虑(gcd(i,j))在(varphi(ij))中的贡献，不难得到(varphi(ij) = frac{varphi(i) varphi(j) gcd(i,j)}{varphi(gcd(i,j))})
代入得$ sumlimits_{i=1}^n sumlimits_{j=1}^n varphi(ij) =sumlimits_{i=1}^n sumlimits_{j=1}^n frac{varphi(i) varphi(j) gcd(i,j)}{varphi(gcd(i,j))}$
按照套路枚举(gcd)：(=sumlimits_{d=1}^n frac{d}{varphi(d)} sumlimits_{i=1}^{lfloor frac{n}{d} floor}sumlimits_{j=1}^{lfloor frac{n}{d} floor} varphi(id) varphi(jd) [gcd(i,j) == 1]=sumlimits_{d=1}^n frac{d}{varphi(d)} sumlimits_{i=1}^{lfloor frac{n}{d} floor}sumlimits_{j=1}^{lfloor frac{n}{d} floor} varphi(id) varphi(jd) sumlimits_{p | gcd(i,j)} mu(p))
将(p)移到前面：(=sumlimits_{d=1}^n sumlimits_{p=1}^{lfloor frac{n}{d} floor} frac{d mu(p)}{varphi(d)} sumlimits_{i=1}^{lfloor frac{n}{dp} floor}sumlimits_{j=1}^{lfloor frac{n}{dp} floor} varphi(idp) varphi(jdp))
(dp)太难看了考虑枚举(T=dp)：(=sumlimits_{T=1}^n sumlimits_{d | T} frac{d mu(frac{T}{d})}{varphi(d)} sumlimits_{i=1}^{lfloor frac{n}{T} floor}sumlimits_{j=1}^{lfloor frac{n}{T} floor} varphi(iT) varphi(jT))
然后将(dist(t_i,t_j))代回来。注意这个时候(i,j)的意义发生了变化，我们应该要代入的是(dist(t_{iT} , t_{jT}))
所以我们需要求的是(=sumlimits_{T=1}^n sumlimits_{d | T} frac{d mu(frac{T}{d})}{varphi(d)} sumlimits_{i=1}^{lfloor frac{n}{T} floor}sumlimits_{j=1}^{lfloor frac{n}{T} floor} varphi(iT) varphi(jT) dist(t_{iT} , t_{jT}))
推到这里我们就可以做了，相对来说还是比较好推的……
对于(sumlimits_{d | T} frac{d mu(frac{T}{d})}{varphi(d)})，枚举倍数做到(O(nlogn))预处理
对于(sumlimits_{i=1}^{lfloor frac{n}{T} floor}sumlimits_{j=1}^{lfloor frac{n}{T} floor} varphi(iT) varphi(jT) dist(t_{iT} , t_{jT}))，它等于(sumlimits_{i=1}^{lfloor frac{n}{T} floor}sumlimits_{j=1}^{lfloor frac{n}{T} floor} varphi(iT) varphi(jT) (dep_{t_{iT}} + dep_{t_{jT}} - 2 * dep_{LCA(t_{iT} , t_{jT})}))。我们把所有点权为(T)的倍数的点拿出来建立虚树进行树形DP，对于每一个点维护它子树中满足条件的点的(sum varphi(i) imes dep_i)与(sum varphi(i))，在两个子树合并时计算答案即可。总复杂度约为(O(nlog^2n))。
最后记录一些被坑的细节（大概只有我这种菜鸡才会犯……）
(1.)建立虚树的点不是编号为(T)的倍数的点，而是点权是(T)的倍数的点
(2.)(frac{d}{varphi(d)})不一定是整数
(3.)写程序的时候要保持清醒……发现很多地方(i)打成(j)，(a_x)打成(x)之类的……
(4.)记得清空数组
#include<bits/stdc++.h>
//This code is written by Itst
using namespace std;
inline int read(){
int a = 0;
char c = getchar();
bool f = 0;
while(!isdigit(c) && c != EOF){
if(c == '-')
f = 1;
c = getchar();
}
if(c == EOF)
exit(0);
while(isdigit(c)){
a = a * 10 + c - 48;
c = getchar();
}
return f ? -a : a;
}
const int MAXN = 2e5 + 9 , MOD = 1e9 + 7;
int mu[MAXN] , phi[MAXN] , prime[MAXN] , cnt;
bool nprime[MAXN];
struct Edge{
int end , upEd;
}Ed[MAXN << 1];
int N , cntEd , ts , top , cntT , cntST , cur , ans1 , ans;
int head[MAXN] , val[MAXN] , ind[MAXN];
int dfn[MAXN] , dep[MAXN] , fir[MAXN] , ST[21][MAXN << 1] , logg2[MAXN << 1];
int st[MAXN] , tree[MAXN] , dp1[MAXN] , dp2[MAXN] , q[MAXN];
vector < int > ch[MAXN];
inline void addEd(int a , int b){
Ed[++cntEd].end = b;
Ed[cntEd].upEd = head[a];
head[a] = cntEd;
}
void input(){
N = read();
for(int i = 1 ; i <= N ; ++i)
ind[val[i] = read()] = i;
for(int i = 1 ; i < N ; ++i){
int a = read() , b = read();
addEd(a , b);
addEd(b , a);
}
}
inline void init(){
phi[1] = mu[1] = 1;
for(int i = 2 ; i <= N ; ++i){
if(!nprime[i]){
prime[++cnt] = i;
phi[i] = i - 1;
mu[i] = -1;
}
for(int j = 1 ; j <= cnt && prime[j] * i <= N ; ++j){
nprime[i * prime[j]] = 1;
if(i % prime[j] == 0){
phi[i * prime[j]] = phi[i] * prime[j];
break;
}
phi[i * prime[j]] = phi[i] * (prime[j] - 1);
mu[i * prime[j]] = mu[i] * -1;
}
}
}
void init_dfs(int x , int p){
dfn[x] = ++ts;
ST[0][++cntST] = x;
fir[x] = cntST;
dep[x] = dep[p] + 1;
for(int i = head[x] ; i ; i = Ed[i].upEd)
if(Ed[i].end != p){
init_dfs(Ed[i].end , x);
ST[0][++cntST] = x;
}
}
inline int cmp(int a , int b){
return dep[a] < dep[b] ? a : b;
}
void init_ST(){
for(int i = 2 ; i <= cntST ; ++i)
logg2[i] = logg2[i >> 1] + 1;
for(int i = 1 ; 1 << i <= cntST ; ++i)
for(int j = 1 ; j + (1 << i) <= cntST + 1 ; ++j)
ST[i][j] = cmp(ST[i - 1][j] , ST[i - 1][j + (1 << (i - 1))]);
}
inline int LCA(int x , int y){
x = fir[x];
y = fir[y];
if(x > y)
swap(x , y);
int t = logg2[y - x + 1];
return cmp(ST[t][x] , ST[t][y - (1 << t) + 1]);
}
inline int poww(long long a , int b){
int times = 1;
while(b){
if(b & 1)
times = times * a % MOD;
a = a * a % MOD;
b >>= 1;
}
return times;
}
bool cmp1(int a , int b){
return dfn[a] < dfn[b];
}
int dfs(int x){
dp1[x] = dp2[x] = 0;
int sum = 0;
for(int i = 0 ; i < ch[x].size() ; ++i){
sum = (sum + dfs(ch[x][i])) % MOD;
sum = (sum + 1ll * dp1[x] * dp2[ch[x][i]] % MOD + 1ll * dp1[ch[x][i]] * dp2[x] % MOD - 2ll * dep[x] * dp1[x] % MOD * dp1[ch[x][i]] % MOD + MOD) % MOD;
dp1[x] = (dp1[x] + dp1[ch[x][i]]) % MOD;
dp2[x] = (dp2[x] + dp2[ch[x][i]]) % MOD;
}
if(val[x] % cur == 0){
sum = (sum + 1ll * dp2[x] * phi[val[x]] % MOD - 1ll * dep[x] * dp1[x] % MOD * phi[val[x]] % MOD + MOD) % MOD;
dp1[x] = (dp1[x] + phi[val[x]]) % MOD;
dp2[x] = (dp2[x] + 1ll * phi[val[x]] * dep[x]) % MOD;
}
return sum;
}
inline void calc(int x){
cntT = 0;
for(int i = 1 ; x * i <= N ; ++i){
ch[ind[x * i]].clear();
tree[++cntT] = ind[x * i];
}
sort(tree + 1 , tree + cntT + 1 , cmp1);
for(int i = 1 ; i <= cntT ; ++i){
if(top){
int t = LCA(st[top] , tree[i]);
while(top - 1 && dep[st[top - 1]] >= dep[t]){
ch[st[top - 1]].push_back(st[top]);
--top;
}
if(t != st[top]){
ch[t].clear();
ch[t].push_back(st[top]);
st[top] = t;
}
}
st[++top] = tree[i];
}
int rt = st[1];
while(top > 1){
ch[st[top - 1]].push_back(st[top]);
--top;
}
top = 0;
ans = (ans + 1ll * q[x] * dfs(rt)) % MOD;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("in","r",stdin);
//freopen("out","w",stdout);
#endif
input();
init();
init_dfs(1 , 0);
init_ST();
for(int i = 1 ; i <= N ; ++i)
for(int j = 1 ; j * i <= N ; ++j)
q[i * j] = (q[i * j] + 1ll * i * mu[j] * poww(phi[i] , MOD - 2) % MOD + MOD) % MOD;
for(cur = 1 ; cur <= N / 2 ; ++cur)
calc(cur);
cout << 2ll * ans * poww(1ll * N * (N - 1) % MOD , MOD - 2) % MOD;
return 0;
}