$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^n\varphi (a_i\times a_j)d(i,j) \\ 设p_{a_i}=i \\ \sum _{i=1}^n\sum _{j=1}^n\varphi (i\times j)d(p_i,p_j) \\ \sum _{i=1}^n\sum _{j=1}^n\frac{\varphi (i)\varphi (j)}{\varphi (\gcd (i,j))}\times \gcd (i,j)\times d(p_i,p_j) \\ \sum _{d=1}^n\frac{d}{\varphi (d)}\sum _{i=1}^{\frac{n}{d}}\sum _{j=1}^{\frac{n}{d}}\varphi (id)\varphi (jd)\times d(p_{id},p_{jd})[\gcd (i,j)=1] \\ \sum _{d=1}^n\frac{d}{\varphi (d)}\sum _{k=1}^{\frac{n}{d}}\mu (k)\sum _{i=1}^{\frac{n}{kd}}\sum _{j=1}^{\frac{n}{kd}}\varphi (ikd)\varphi (jkd)\times d(p_{ikd},p_{jkd}) \\ T=kd \\ \sum _{T=1}^n\left(\sum _{i=1}^{\frac{n}{T}}\sum _{j=1}^{\frac{n}{T}}\varphi (iT)\varphi (jT)\times d(p_{iT},p_{jT})\right)\sum _{d\mid T}\frac{d}{\varphi (d)}\times \mu (\frac{T}{d}) \end{gathered}$$

我们设$h(n)=\sum _{d\mid n}\frac{d}{\varphi (d)}\times \mu (\frac{n}{d})$，应该是个积性函数吧，但是$O(n\log n)$方便，，，

考虑如何求解$F(T)=\sum _{i=1}^{\frac{n}{T}}\sum _{j=1}^{\frac{n}{T}}\varphi (iT)\varphi (jT)\times d(p_{iT},p_{jT})$，为了方便，简化为$T=1$的情况。

$\sum _{i=1}^n\sum _{j=1}^n\varphi (i)\varphi (j)\times d(p_i,p_j)=\sum _{i=1}^n\varphi (i)\sum _{j=1}^n\varphi (j)\times d(p_i,p_j)$。

考虑信息在两棵子树上合并，$\sum _{i\in U}w(i)\sum _{j\in V}w(j)\times d(i\times j)=\sum _{i\in U}w(i)\sum _{j\in V}w(j)\times (d(i,u)+len+d(v,j))$。
$$\sum _{i\in U}w(i)\times d(i,u)\sum _{j\in V}w(j)+len\sum _{i\in U}w(i)\sum _{j\in V}w(j)+\sum _{i\in U}w(i)\sum _{j\in V}w(j)\times d(v,j)$$
于是设$f(x)=\sum _{i\in X}w(i)\times d(i,x),g(x)=\sum _{j\in X}w(j)$，于是有上式子为$f(U)\times g(V)+len\times g(U)\times g(v)+g(U)\times f(V)$。

同样地，对于$f(x),g(x)$的转移也比较简单，$f(U)=f(U)+len\times g(V)+f(V),g(U)=g(U)+g(V)$。

那么$F(U)=F(U)+F(V)+\left(f(U)\times g(V)+len\times g(U)\times g(v)+g(U)\times f(V)\right)$，最后即可得到我们所要的答案。

上面的计算可以用虚树求解，这道题的整体复杂度$O(n{\log }^{2}n)$。

#include <bits/stdc++.h>using namespace std;const int N = 2e5 + 10, mod = 1e9 + 7;int head[N], to[N << 1], nex[N << 1], cnt = 1;int son[N], sz[N], fa[N], dep[N], id[N], top[N], tot;int n, a[N], p[N], stk[N], prime[N], phi[N], mu[N], h[N], inv[N], num, tp;int f[N], g[N], F[N], w[N];vector< pair<int, int> > G[N];bool st[N];inline int Add(int x, int y) {return x + y < mod ? x + y : x + y - mod;
}inline int Sub(int x, int y) {return x >= y ? x - y : x - y + mod;
}void init() {phi[1] = mu[1] = inv[1] = 1;for (int i = 2; i < N; i++) {inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;if (!st[i]) {prime[++num] = i;mu[i] = -1;phi[i] = i - 1;}for (int j = 1; j <= num && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if (i % prime[j] == 0) {phi[i * prime[j]] = phi[i] * prime[j];break;}mu[i * prime[j]] = -mu[i];phi[i * prime[j]] = phi[i] * (prime[j] - 1);}}for (int i = 1; i < N; i++) {for (int j = i; j < N; j += i) {if (mu[j / i]) {if (mu[j / i] == -1) {h[j] = Sub(h[j], 1ll * i * inv[phi[i]] % mod);}else {h[j] = Add(h[j], 1ll * i * inv[phi[i]] % mod);}}}}
}void add(int x, int y) {to[cnt] = y;nex[cnt] = head[x];head[x] = cnt++;
}int lca(int u, int v) {while (top[u] != top[v]) {if (dep[top[u]] < dep[top[v]]) {swap(u, v);}u = fa[top[u]];}return dep[u] < dep[v] ? u : v;
}void dfs1(int rt, int f) {sz[rt] = 1, dep[rt] = dep[f] + 1, fa[rt] = f, id[rt] = ++tot;for (int i = head[rt]; i; i = nex[i]) {if (to[i] == f) {continue;}dfs1(to[i], rt);sz[rt] += sz[to[i]];if (!son[rt] || sz[to[i]] > sz[son[rt]]) {son[rt] = to[i];}}
}void dfs2(int rt, int tp) {top[rt] = tp;if (!son[rt]) {return; }dfs2(son[rt], tp);for (int i = head[rt]; i; i = nex[i]) {if (to[i] == fa[rt] || to[i] == son[rt]) {continue;}dfs2(to[i], to[i]);}
}bool cmp(int a, int b) {return id[a] < id[b];
}void dp(int u, int fa) {f[u] = 0, g[u] = w[u], F[u] = 0;for (auto to : G[u]) {int len = to.second, v = to.first;if (v == fa) {continue;}dp(v, u);F[u] = Add(Add(F[u], F[v]), Add(Add(1ll * f[u] * g[v] % mod, 1ll * len * g[u] % mod * g[v] % mod), 1ll * g[u] * f[v] % mod));f[u] = Add(Add(f[u], f[v]), 1ll * len * g[v] % mod);g[u] = Add(g[u], g[v]);}G[u].clear();
}void insert(int rt) {if (tp == 1) {if (rt != 1) {stk[++tp] = rt;}return ;}int lc = lca(rt, stk[tp]);if (lc == stk[tp]) {stk[++tp] = rt;return ;}while (tp > 1 && id[stk[tp - 1]] >= id[lc]) {int u = stk[tp - 1], v = stk[tp];G[u].push_back({v, dep[v] - dep[u]});tp--;}if (stk[tp] != lc) {int u = lc, v = stk[tp];G[u].push_back({v, dep[v] - dep[u]});stk[tp] = lc;}stk[++tp] = rt;
}int calc(int T) {int tot = 0;for (int i = T; i <= n; i += T) {a[++tot] = p[i];w[p[i]] = phi[i];}sort(a + 1, a + 1 + tot, cmp);stk[tp = 1] = 1;for (int i = 1; i <= tot; i++) {insert(a[i]);}while (tp > 1) {int u = stk[tp - 1], v = stk[tp];G[u].push_back({v, dep[v] - dep[u]});tp--;}dp(1, 0);for (int i = 1; i <= tot; i++) {w[a[i]] = 0;}return F[1];
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);init();scanf("%d", &n);for (int i = 1; i <= n; i++) {scanf("%d", &a[i]);p[a[i]] = i;}for (int i = 1, u, v; i < n; i++) {scanf("%d %d", &u, &v);add(u, v);add(v, u);}dfs1(1, 0);dfs2(1, 1);int ans = 0;for (int i = 1; i <= n; i++) {ans = Add(ans, 1ll * calc(i) * h[i] % mod);}printf("%lld\n", 1ll * ans * inv[n] % mod * inv[n - 1] % mod * 2 % mod);return 0;
}