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 long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
inline long long power(long long a, long long b) {
  long long ans = 1;
  for (; b; b >>= 1) {
    if (b & 1) ans = ans * a % 1000000007;
    a = a * a % 1000000007;
  }
  return ans;
}
inline long long inv(long long x) { return power(x, 1000000007 - 2); }
struct node {
  long long to, next;
};
struct Tree {
  node e[1010 << 1];
  long long head[1010], size[1010], bit[1010], son[1010][1010], n, tot;
  inline void add(long long x, long long y) {
    e[++tot].to = y;
    e[tot].next = head[x];
    head[x] = tot;
  }
  inline void read1() {
    n = read();
    for (long long i = 1; i < n; i++) {
      long long x = read(), y = read();
      add(x, y);
      add(y, x);
    }
  }
  inline void dfs(long long x, long long fa) {
    size[x] = bit[x] = 0;
    for (long long i = head[x]; i; i = e[i].next)
      if (e[i].to ^ fa)
        dfs(son[x][++size[x]] = e[i].to, x), bit[x] |= 1 << e[i].to - 1;
  }
} S, T;
long long dp[1010][1 << 12], ans, ans1;
bool vis[1010][1 << 12];
inline long long dfs(long long f, long long y, long long qwq) {
  if (!y) return !qwq;
  long long x = S.son[f][y];
  if (vis[x][qwq]) return dp[x][qwq];
  vis[x][qwq] = true;
  long long &res = dp[x][qwq];
  res = dfs(f, y - 1, qwq);
  for (long long i = 0; i < T.n; i++)
    if (qwq >> i & 1)
      res = (dfs(f, y - 1, qwq ^ (1 << i)) * dfs(x, S.size[x], T.bit[i + 1]) +
             res) %
            1000000007;
  return res;
}
signed main() {
  S.read1();
  T.read1();
  S.dfs(1, 0);
  for (long long i = 1; i <= T.n; i++) {
    memset(dp, 0, sizeof dp);
    memset(vis, false, sizeof vis);
    T.dfs(i, 0);
    for (long long j = 1; j <= S.n; j++)
      (ans += dfs(j, S.size[j], T.bit[i])) %= 1000000007;
  }
  S = T;
  S.dfs(1, 0);
  for (long long i = 1; i <= T.n; i++) {
    memset(dp, 0, sizeof dp);
    memset(vis, false, sizeof vis);
    T.dfs(i, 0);
    (ans1 += dfs(1, S.size[1], T.bit[i])) %= 1000000007;
  }
  ans = ans * inv(ans1) % 1000000007;
  return printf("%lld\n", ans), 0;
}