Problem: Let the main characters of this problem be personages from some recent movie. New Avengers seem to make a lot of buzz. I didn't watch any part of the franchise and don't know its heroes well, but it won't stop me from using them in this problem statement. So, Thanos and Dr. Strange are doing their superhero and supervillain stuff, but then suddenly they stumble across a regular competitive programming problem.

You are given a tree with n vertices.

In each vertex v there is positive integer a_{v}.

You have to answer q queries.

Each query has a from u v x.

You have to calculate ∏_{w ∈ P} gcd(x, a_{w}) mod (10^{9} + 7), where P is a set of vertices on path from u to v. In other words, you are to calculate the product of gcd(x, a_{w}) for all vertices w on the path from u to v. As it might be large, compute it modulo 10^9+7. Here gcd(s, t) denotes the greatest common divisor of s and t.

Note that the numbers in vertices do not change after queries.

I suppose that you are more interested in superhero business of Thanos and Dr. Strange than in them solving the problem. So you are invited to solve this problem instead of them.

Input

In the first line of input there is one integer n (1 ≤ n ≤ 10^{5}) — the size of the tree.

In the next n-1 lines the edges of the tree are described. The i-th edge is described with two integers u_{i} and v_{i} (1 ≤ u_{i}, v_{i} ≤ n) and it connects the vertices u_{i} and v_{i}. It is guaranteed that graph with these edges is a tree.

In the next line there are n integers a_1, a_2, …, a_n (1 ≤ a_{v} ≤ 10^{7}).

In the next line there is one integer q (1 ≤ q ≤ 10^{5}) — the number of queries.

And in the next q lines the queries are described. Each query is described with three integers u_{i}, v_{i} and x_{i} (1 ≤ u_{i}, v_{i} ≤ n, 1 ≤ x_{i} ≤ 10^{7}).

Output

Print q numbers — the answers to the queries in the order they are given in the input. Print each answer modulo 10^9+7 = 1000000007. Print each number on a separate line.

Examples

Input

4
1 2
1 3
1 4
6 4 9 5
3
2 3 6
2 3 2
3 4 7


Output

36
4
1


Input

6
1 2
2 3
2 4
1 5
5 6
100000 200000 500000 40000 800000 250000
3
3 5 10000000
6 2 3500000
4 1 64000


Output

196000
12250
999998215

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char *s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
const int md = (int)1e9 + 7;
inline void add(int &a, int b) {
  a += b;
  if (a >= md) a -= md;
}
inline void sub(int &a, int b) {
  a -= b;
  if (a < 0) a += md;
}
inline int mul(int a, int b) {
  return (int)((long long)a * b % md);
  unsigned long long x = (long long)a * b;
  unsigned xh = (unsigned)(x >> 32), xl = (unsigned)x, d, m;
  asm("divl %4; \n\t" : "=a"(d), "=d"(m) : "d"(xh), "a"(xl), "r"(md));
  return m;
}
inline int power(int a, long long b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = mul(res, a);
    }
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
inline int inv(int a) {
  a %= md;
  if (a < 0) a += md;
  int b = md, u = 0, v = 1;
  while (a) {
    int t = b / a;
    b -= t * a;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  assert(b == 1);
  if (u < 0) u += md;
  return u;
}
template <typename T>
class graph {
 public:
  struct edge {
    int from;
    int to;
    T cost;
  };
  vector<edge> edges;
  vector<vector<int>> g;
  int n;
  function<bool(int)> ignore;
  graph(int _n) : n(_n) {
    g.resize(n);
    ignore = nullptr;
  }
  virtual int add(int from, int to, T cost) = 0;
  virtual void set_ignore_edge_rule(const function<bool(int)> &f) {
    ignore = f;
  }
  virtual void reset_ignore_edge_rule() { ignore = nullptr; }
};
template <typename T>
class forest : public graph<T> {
 public:
  using graph<T>::edges;
  using graph<T>::g;
  using graph<T>::n;
  forest(int _n) : graph<T>(_n) {}
  int add(int from, int to, T cost = 1) {
    assert(0 <= from && from < n && 0 <= to && to < n);
    int id = (int)edges.size();
    assert(id < n - 1);
    g[from].push_back(id);
    g[to].push_back(id);
    edges.push_back({from, to, cost});
    return id;
  }
};
template <typename T>
class dfs_forest : public forest<T> {
 public:
  using forest<T>::edges;
  using forest<T>::g;
  using forest<T>::n;
  using forest<T>::ignore;
  vector<int> pv;
  vector<int> pe;
  vector<int> order;
  vector<int> pos;
  vector<int> end;
  vector<int> sz;
  vector<int> root;
  vector<int> depth;
  vector<T> dist;
  dfs_forest(int _n) : forest<T>(_n) {}
  void init() {
    pv = vector<int>(n, -1);
    pe = vector<int>(n, -1);
    order.clear();
    pos = vector<int>(n, -1);
    end = vector<int>(n, -1);
    sz = vector<int>(n, 0);
    root = vector<int>(n, -1);
    depth = vector<int>(n, -1);
    dist = vector<T>(n);
  }
  void clear() {
    pv.clear();
    pe.clear();
    order.clear();
    pos.clear();
    end.clear();
    sz.clear();
    root.clear();
    depth.clear();
    dist.clear();
  }

 private:
  void do_dfs(int v) {
    pos[v] = (int)order.size();
    order.push_back(v);
    sz[v] = 1;
    for (int id : g[v]) {
      if (id == pe[v] || (ignore != nullptr && ignore(id))) {
        continue;
      }
      auto &e = edges[id];
      int to = e.from ^ e.to ^ v;
      depth[to] = depth[v] + 1;
      dist[to] = dist[v] + e.cost;
      pv[to] = v;
      pe[to] = id;
      root[to] = (root[v] != -1 ? root[v] : to);
      do_dfs(to);
      sz[v] += sz[to];
    }
    end[v] = (int)order.size() - 1;
  }
  void do_dfs_from(int v) {
    depth[v] = 0;
    dist[v] = T{};
    root[v] = v;
    pv[v] = pe[v] = -1;
    do_dfs(v);
  }

 public:
  void dfs(int v, bool clear_order = true) {
    if (pv.empty()) {
      init();
    } else {
      if (clear_order) {
        order.clear();
      }
    }
    do_dfs_from(v);
  }
  void dfs_all() {
    init();
    for (int v = 0; v < n; v++) {
      if (depth[v] == -1) {
        do_dfs_from(v);
      }
    }
    assert((int)order.size() == n);
  }
};
template <typename T>
class lca_forest : public dfs_forest<T> {
 public:
  using dfs_forest<T>::edges;
  using dfs_forest<T>::g;
  using dfs_forest<T>::n;
  using dfs_forest<T>::pv;
  using dfs_forest<T>::pos;
  using dfs_forest<T>::end;
  using dfs_forest<T>::depth;
  int h;
  vector<vector<int>> pr;
  lca_forest(int _n) : dfs_forest<T>(_n) {}
  inline void build_lca() {
    assert(!pv.empty());
    int max_depth = 0;
    for (int i = 0; i < n; i++) {
      max_depth = max(max_depth, depth[i]);
    }
    h = 1;
    while ((1 << h) <= max_depth) {
      h++;
    }
    pr.resize(n);
    for (int i = 0; i < n; i++) {
      pr[i].resize(h);
      pr[i][0] = pv[i];
    }
    for (int j = 1; j < h; j++) {
      for (int i = 0; i < n; i++) {
        pr[i][j] = (pr[i][j - 1] == -1 ? -1 : pr[pr[i][j - 1]][j - 1]);
      }
    }
  }
  inline bool anc(int x, int y) {
    return (pos[x] <= pos[y] && end[y] <= end[x]);
  }
  inline int lca(int x, int y) {
    assert(!pr.empty());
    if (anc(x, y)) {
      return x;
    }
    if (anc(y, x)) {
      return y;
    }
    for (int j = h - 1; j >= 0; j--) {
      if (pr[x][j] != -1 && !anc(pr[x][j], y)) {
        x = pr[x][j];
      }
    }
    return pr[x][0];
  }
};
template <typename T>
class fenwick {
 public:
  vector<T> fenw;
  int n;
  fenwick(int _n) : n(_n) { fenw.resize(n); }
  void modify(int x, T v) {
    while (x < n) {
      fenw[x] += v;
      x |= (x + 1);
    }
  }
  T get(int x) {
    T v{};
    while (x >= 0) {
      v += fenw[x];
      x = (x & (x + 1)) - 1;
    }
    return v;
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  const int MAX = (int)1e7 + 10;
  vector<int> prime_div(MAX);
  for (int i = 0; i < MAX; i++) {
    prime_div[i] = i;
  }
  for (int i = 2; i * i < MAX; i++) {
    if (prime_div[i] == i) {
      for (int j = i * i; j < MAX; j += i) {
        if (prime_div[j] == j) {
          prime_div[j] = i;
        }
      }
    }
  }
  vector<int> pos(MAX);
  vector<int> primes;
  for (int i = 2; i < MAX; i++) {
    if (prime_div[i] == i) {
      pos[i] = (int)primes.size();
      primes.push_back(i);
    }
  }
  vector<vector<pair<int, int>>> at(primes.size());
  int n;
  cin >> n;
  lca_forest<int> g(n);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    g.add(x, y);
  }
  g.dfs(0);
  g.build_lca();
  for (int i = 0; i < n; i++) {
    int foo;
    cin >> foo;
    while (foo > 1) {
      int d = prime_div[foo];
      int cc = 0;
      while (foo % d == 0) {
        foo /= d;
        cc++;
      }
      at[pos[d]].emplace_back(cc, i);
    }
  }
  int tt;
  cin >> tt;
  vector<int> u(tt), v(tt), lca(tt);
  vector<int> res(tt, 1);
  for (int i = 0; i < tt; i++) {
    int foo;
    cin >> u[i] >> v[i] >> foo;
    u[i]--;
    v[i]--;
    lca[i] = g.lca(u[i], v[i]);
    while (foo > 1) {
      int d = prime_div[foo];
      int cc = 0;
      while (foo % d == 0) {
        foo /= d;
        cc++;
      }
      for (int j = 1; j <= cc; j++) {
        at[pos[d]].emplace_back(j, ~i);
      }
    }
  }
  fenwick<int> fenw(n);
  for (int it = 0; it < (int)primes.size(); it++) {
    if (at[it].empty()) {
      continue;
    }
    sort(at[it].rbegin(), at[it].rend());
    for (auto &p : at[it]) {
      int id = p.second;
      if (id >= 0) {
        fenw.modify(g.pos[id], 1);
        fenw.modify(g.end[id] + 1, -1);
      } else {
        id = ~id;
        int cnt = fenw.get(g.pos[u[id]]) + fenw.get(g.pos[v[id]]) -
                  fenw.get(g.pos[lca[id]]);
        if (lca[id] != 0) {
          cnt -= fenw.get(g.pos[g.pv[lca[id]]]);
        }
        res[id] = mul(res[id], power(primes[it], cnt));
      }
    }
    for (auto &p : at[it]) {
      int id = p.second;
      if (id >= 0) {
        fenw.modify(g.pos[id], -1);
        fenw.modify(g.end[id] + 1, +1);
      }
    }
  }
  for (int i = 0; i < tt; i++) {
    cout << res[i] << '\n';
  }
  return 0;
}