Problem: Tired of boring dates, Leha and Noora decided to play a game.

Leha found a tree with n vertices numbered from 1 to n. We remind you that tree is an undirected graph without cycles. Each vertex v of a tree has a number av written on it. Quite by accident it turned out that all values written on vertices are distinct and are natural numbers between 1 and n.

The game goes in the following way. Noora chooses some vertex u of a tree uniformly at random and passes a move to Leha. Leha, in his turn, chooses (also uniformly at random) some vertex v from remaining vertices of a tree (v ≠ u). As you could guess there are n(n - 1) variants of choosing vertices by players. After that players calculate the value of a function f(u, v) = φ(au·av) · d(u, v) of the chosen vertices where φ(x) is Euler's totient function and d(x, y) is the shortest distance between vertices x and y in a tree.

Soon the game became boring for Noora, so Leha decided to defuse the situation and calculate expected value of function f over all variants of choosing vertices u and v, hoping of at least somehow surprise the girl.

Leha asks for your help in calculating this expected value. Let this value be representable in the form of an irreducible fraction <image>. To further surprise Noora, he wants to name her the value <image>. 

Help Leha!

Input

The first line of input contains one integer number n (2 ≤ n ≤ 2·105) — number of vertices in a tree.

The second line contains n different numbers a1, a2, ..., an (1 ≤ ai ≤ n) separated by spaces, denoting the values written on a tree vertices.

Each of the next n - 1 lines contains two integer numbers x and y (1 ≤ x, y ≤ n), describing the next edge of a tree. It is guaranteed that this set of edges describes a tree.

Output

In a single line print a number equal to P·Q - 1 modulo 109 + 7.

Examples

Input

3
1 2 3
1 2
2 3


Output

333333338


Input

5
5 4 3 1 2
3 5
1 2
4 3
2 5


Output

8

Note

Euler's totient function φ(n) is the number of such i that 1 ≤ i ≤ n,and gcd(i, n) = 1, where gcd(x, y) is the greatest common divisor of numbers x and y.

There are 6 variants of choosing vertices by Leha and Noora in the first testcase:

  * u = 1, v = 2, f(1, 2) = φ(a1·a2)·d(1, 2) = φ(1·2)·1 = φ(2) = 1
  * u = 2, v = 1, f(2, 1) = f(1, 2) = 1
  * u = 1, v = 3, f(1, 3) = φ(a1·a3)·d(1, 3) = φ(1·3)·2 = 2φ(3) = 4
  * u = 3, v = 1, f(3, 1) = f(1, 3) = 4
  * u = 2, v = 3, f(2, 3) = φ(a2·a3)·d(2, 3) = φ(2·3)·1 = φ(6) = 2
  * u = 3, v = 2, f(3, 2) = f(2, 3) = 2



Expected value equals to <image>. The value Leha wants to name Noora is 7·3 - 1 = 7·333333336 = 333333338 <image>.

In the second testcase expected value equals to <image>, so Leha will have to surprise Hoora by number 8·1 - 1 = 8 <image>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int inf = (1ll << 30) - 1;
const int maxn = (int)1e5 + 10;
using namespace std;
int n;
int a[200200];
vector<int> g[200200];
vector<int> pr[200200];
int P[200200][70];
int Q[200200][70];
int inv[200200][70];
int sz[200200];
long long cur[200200];
long long cur2[200200];
void dfs(int v, int p) {
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to == p) continue;
    dfs(to, v);
    sz[v] += sz[to];
  }
  ++sz[v];
}
int mod = 1000 * 1000 * 1000 + 7;
int p = 0;
int pc[511];
long long ansA = 0;
long long ansB = 0;
int SH = 0;
void add(int x, long long deg) {
  int sz = pr[x].size();
  for (int i = 0; i < (1 << sz); i++) {
    int D = Q[x][i];
    cur[D] = (cur[D] + (deg - SH + mod * 1ll) % mod * 1ll *
                           (x / Q[x][(1 << sz) - 1]) % mod *
                           P[x][(1 << sz) - 1] % mod) %
             mod;
    cur2[D] = (cur2[D] +
               (x / Q[x][(1 << sz) - 1]) % mod * P[x][(1 << sz) - 1] % mod) %
              mod;
  }
}
void del(int x) {
  int sz = pr[x].size();
  for (int i = 0; i < (1 << sz); i++) {
    int D = Q[x][i];
    cur[D] = 0;
    cur2[D] = 0;
  }
}
void S(int v, int p, int len) {
  add(a[v], len);
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to == p) continue;
    S(to, v, len + 1);
  }
}
long long tmp[1111];
long long tmp2[1111];
void D(int v, int p) {
  del(a[v]);
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to == p) continue;
    D(to, v);
  }
}
void comp(int v, int len) {
  int x = a[v];
  int sz = pr[x].size();
  long long DD = (x / Q[x][(1 << sz) - 1]) * 1ll * P[x][(1 << sz) - 1];
  for (int i = 0; i < (1 << sz); i++) tmp[i] = tmp2[i] = 0;
  for (int mask = (1 << sz) - 1; mask >= 0; --mask) {
    for (int mask2 = mask;; mask2 = (mask2 - 1) & mask) {
      int pp = Q[x][mask];
      long long &A = tmp[mask2];
      long long &B = tmp2[mask2];
      long long D = (cur[pp] + SH * cur2[pp] % mod) % mod;
      if (pc[mask2 ^ mask] & 1) {
        A -= D;
        if (A < 0) A += mod;
        B -= cur2[pp] % mod;
        if (B < 0) B += mod;
      } else {
        A += D;
        if (A >= mod) A -= mod;
        B += cur2[pp] % mod;
        if (B >= mod) B -= mod;
      }
      if (mask2 == 0) break;
    }
    ansA = (ansA + (tmp[mask] + tmp2[mask] * len % mod) % mod * Q[x][mask] %
                       mod * DD % mod * inv[x][mask] % mod) %
           mod;
  }
}
void calc2(int v, int p, int len) {
  comp(v, len);
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to == p) continue;
    calc2(to, v, len + 1);
  }
}
void calc(int v, int p, int second) {
  int mx = -1;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to == p) continue;
    if (mx == -1 || sz[mx] < sz[to]) mx = to;
  }
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (to == p || mx == to) continue;
    calc(to, v, 0);
  }
  if (mx != -1) {
    calc(mx, v, 1);
    SH++;
    comp(v, 0);
    add(a[v], 0);
    for (int i = 0; i < g[v].size(); i++) {
      int to = g[v][i];
      if (to == p || mx == to) continue;
      calc2(to, v, 1);
      S(to, v, 1);
    }
  } else {
    add(a[v], 0);
  }
  if (!second) {
    D(v, p);
    SH = 0;
  }
}
long long binpow(long long x, int y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * 1ll * x) % mod;
    x = (x * x) % mod;
    y >>= 1;
  }
  return res;
}
void solve() {
  scanf("%d", &n);
  for (int i = 2; i <= max(n, 1000); i++) {
    if (pr[i].size() == 0) {
      for (int j = i; j <= max(n, 1000); j += i) pr[j].push_back(i);
    }
  }
  for (int i = 1; i <= max(n, 1000); i++) {
    int len = pr[i].size();
    for (int mask = 0; mask < (1 << len); mask++) {
      P[i][mask] = 1;
      Q[i][mask] = 1;
      for (int j = 0; j < len; j++) {
        if (mask & (1 << j)) {
          P[i][mask] *= pr[i][j] - 1;
          Q[i][mask] *= pr[i][j];
        }
      }
      inv[i][mask] = binpow(P[i][mask], mod - 2);
    }
  }
  for (int i = 0; i < 500; i++) pc[i] = __builtin_popcount(i);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  calc(1, 0, 1);
  cout << ansA * binpow((n * 1ll * (n - 1) / 2) % mod, mod - 2) % mod << endl;
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}