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;
inline int read() {
  int w = 0, x = 0;
  char c = getchar();
  while (!isdigit(c)) w |= c == '-', c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return w ? -x : x;
}
namespace star {
const int maxn = 1005, mod = 1e9 + 7, maxf = (1 << 12);
int n, m;
struct tree {
  int n, ecnt, head[maxn], bit[maxn], to[maxn << 1], nxt[maxn << 1];
  vector<int> son[maxn];
  inline void addedge(int a, int b) {
    to[++ecnt] = b, nxt[ecnt] = head[a], head[a] = ecnt;
    to[++ecnt] = a, nxt[ecnt] = head[b], head[b] = ecnt;
  }
  void dfs(int x, int fa) {
    son[x].clear();
    bit[x] = 0;
    for (int i = head[x]; i; i = nxt[i]) {
      int u = to[i];
      if (u == fa) continue;
      son[x].push_back(u);
      dfs(u, x);
      bit[x] |= (1 << (u - 1));
    }
  }
  inline void init() {
    n = read();
    for (int i = 1; i < n; i++) addedge(read(), read());
  }
} S, T;
inline long long fpow(long long a, int b) {
  long long ans = 1;
  for (; b; b >>= 1, a = a * a % mod)
    if (b & 1) ans = ans * a % mod;
  return ans;
}
long long f[maxn][maxf], ans1, ans2;
bool vis[maxn][maxf];
long long dfs(int x, int id, int d) {
  if (!id) return !d;
  int u = S.son[x][id - 1];
  if (vis[u][d]) return f[u][d];
  vis[u][d] = 1;
  long long sum = dfs(x, id - 1, d);
  for (int i = 0; i < T.n; i++)
    if (d >> i & 1)
      sum = (sum + dfs(x, id - 1, d ^ (1 << i)) *
                       dfs(u, S.son[u].size(), T.bit[i + 1]) % mod) %
            mod;
  return f[u][d] = sum;
}
inline void work() {
  S.init(), T.init();
  S.dfs(1, 0);
  for (int i = 1; i <= T.n; i++) {
    memset(vis, 0, sizeof vis);
    memset(f, 0, sizeof f);
    T.dfs(i, 0);
    for (int j = 1; j <= S.n; j++)
      ans1 = (ans1 + dfs(j, S.son[j].size(), T.bit[i])) % mod;
  }
  S = T;
  S.dfs(1, 0);
  for (int i = 1; i <= T.n; i++) {
    memset(vis, 0, sizeof vis);
    memset(f, 0, sizeof f);
    T.dfs(i, 0);
    ans2 = (ans2 + dfs(1, S.son[1].size(), T.bit[i])) % mod;
  }
  printf("%lld\n", ans1 * fpow(ans2, mod - 2) % mod);
}
}  // namespace star
signed main() {
  star::work();
  return 0;
}