Problem: You are given two trees (connected undirected acyclic graphs) S and T.

Count the number of subtrees (connected subgraphs) of S that are isomorphic to tree T. Since this number can get quite large, output it modulo 109 + 7.

Two subtrees of tree S are considered different, if there exists a vertex in S that belongs to exactly one of them.

Tree G is called isomorphic to tree H if there exists a bijection f from the set of vertices of G to the set of vertices of H that has the following property: if there is an edge between vertices A and B in tree G, then there must be an edge between vertices f(A) and f(B) in tree H. And vice versa — if there is an edge between vertices A and B in tree H, there must be an edge between f - 1(A) and f - 1(B) in tree G.

Input

The first line contains a single integer |S| (1 ≤ |S| ≤ 1000) — the number of vertices of tree S.

Next |S| - 1 lines contain two integers ui and vi (1 ≤ ui, vi ≤ |S|) and describe edges of tree S.

The next line contains a single integer |T| (1 ≤ |T| ≤ 12) — the number of vertices of tree T.

Next |T| - 1 lines contain two integers xi and yi (1 ≤ xi, yi ≤ |T|) and describe edges of tree T.

Output

On the first line output a single integer — the answer to the given task modulo 109 + 7.

Examples

Input

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


Output

3


Input

3
2 3
3 1
3
1 2
1 3


Output

1


Input

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


Output

20


Input

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


Output

0

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename tn>
void read(tn &a) {
  tn x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - 48;
  a = x * f;
}
int get(int x, int id) { return (x >> id) & 1; }
const int mod = 1e9 + 7;
int Mod(int x) { return x >= mod ? x - mod : x; }
template <typename tn>
void Inc(tn &x, int y) {
  x = Mod(x + y);
}
map<int, vector<int> > hsh;
map<int, int> g[2050], tree;
vector<int> son[2050], E[2050];
int f[2][1 << 13];
int val[2050], p[2050];
int n, m, Ans;
void dfs(int u, int fa) {
  val[u] = 1;
  vector<int> vec;
  vec.clear();
  for (auto v : E[u])
    if (v != fa) dfs(v, u), vec.push_back(val[v]);
  sort(vec.begin(), vec.end());
  for (int i = 0, si = vec.size(); i < si; i++) val[u] += vec[i] * p[i];
  if (!hsh.count(val[u])) hsh[val[u]] = vec;
}
void solve(int u, int fa) {
  for (auto v : son[u])
    if (v != fa) solve(v, u);
  for (auto p : hsh) {
    auto &vec = p.second;
    int val = p.first;
    memset(f, 0, sizeof f), f[0][0] = 1;
    int si = vec.size(), lst = 0, now = 1;
    for (auto v : son[u])
      if (v != fa) {
        memcpy(f[now], f[lst], sizeof f[lst]);
        for (int i = 0; i < (1 << si); i++) {
          for (int j = 0, k; j < si; j++)
            if (!get(i, j) && g[v][vec[j]]) {
              Inc(f[now][i | (1 << j)], 1ll * f[lst][i] * g[v][vec[j]] % mod);
              for (k = j; k < si && vec[k] == vec[j]; k++)
                ;
              j = k - 1;
            }
        }
        lst = now, now = lst ^ 1;
      }
    (g[u][val] += f[lst][(1 << si) - 1]) %= mod;
    if (tree[val]) Inc(Ans, g[u][val]);
  }
}
bool prime(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return 0;
  return 1;
}
int gprime() {
  int x = rand();
  while (!prime(x)) x = rand();
  return x;
}
void init() {
  srand(time(0));
  for (int i = 0; i <= n; i++) p[i] = gprime();
}
int main() {
  read(n), init();
  for (int i = 1, u, v; i < n; i++)
    read(u), read(v), son[u].push_back(v), son[v].push_back(u);
  read(m);
  for (int i = 1, u, v; i < m; i++)
    read(u), read(v), E[u].push_back(v), E[v].push_back(u);
  for (int i = 1; i <= m; i++) dfs(i, 0), tree[val[i]] = true;
  solve(1, 0);
  cout << Ans << '\n';
  return 0;
}