You are given a tree consisting of n vertices. A number is written on each vertex; the number on vertex i is equal to a_i.

Let's denote the function g(x, y) as the greatest common divisor of the numbers written on the vertices belonging to the simple path from vertex x to vertex y (including these two vertices). Also let's denote dist(x, y) as the number of vertices on the simple path between vertices x and y, including the endpoints. dist(x, x) = 1 for every vertex x.

Your task is calculate the maximum value of dist(x, y) among such pairs of vertices that g(x, y) > 1.

Input

The first line contains one integer n — the number of vertices (1 ≤ n ≤ 2 ⋅ 10^5).

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 2 ⋅ 10^5) — the numbers written on vertices.

Then n - 1 lines follow, each containing two integers x and y (1 ≤ x, y ≤ n, x ≠ y) denoting an edge connecting vertex x with vertex y. It is guaranteed that these edges form a tree.

Output

If there is no pair of vertices x, y such that g(x, y) > 1, print 0. Otherwise print the maximum value of dist(x, y) among such pairs.

Examples

Input

3
2 3 4
1 2
2 3


Output

1


Input

3
2 3 4
1 3
2 3


Output

2


Input

3
1 1 1
1 2
2 3


Output

0
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
struct fk {
  int to, nxt;
} e[N << 1];
int n, head[N], ans = 1, cnt, flag = 1;
vector<int> p[N], f[N];
int read() {
  char ch = getchar();
  int x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x * f;
}
void add(int u, int v) {
  e[++cnt].to = v;
  e[cnt].nxt = head[u];
  head[u] = cnt;
}
void upd(int u, int v) {
  for (int i = 0; i < p[u].size(); i++)
    for (int j = 0; j < p[v].size(); j++)
      if (p[u][i] == p[v][j]) {
        ans = max(ans, f[u][i] + f[v][j]);
        f[u][i] = max(f[u][i], f[v][j] + 1);
      }
}
void dfs(int u, int fa) {
  for (int v, i = head[u]; i; i = e[i].nxt)
    if ((v = e[i].to) != fa) dfs(v, u), upd(u, v);
}
void divide(int id, int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) {
      p[id].push_back(i), f[id].push_back(1);
      while (x % i == 0) x /= i;
    }
  if (x > 1) p[id].push_back(x), f[id].push_back(1);
}
int main() {
  n = read();
  for (int i = 1, x; i <= n; i++) {
    divide(i, x = read());
    if (x != 1) flag = 0;
  }
  if (flag) {
    puts("0");
    return 0;
  }
  for (int i = 1, u, v; i < n; i++)
    u = read(), v = read(), add(u, v), add(v, u);
  dfs(1, 0);
  printf("%d", ans);
}
