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>
const int N = 2005, M = 13, mod = 1e9 + 7;
using namespace std;
inline int Getint() {
  register int x = 0, f = 1;
  register char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, m, h[N];
struct Edge {
  int to, next;
} g[N << 1];
void AddEdge(int x, int y) {
  static int cnt = 0;
  g[++cnt].to = y, g[cnt].next = h[x], h[x] = cnt;
}
int rd[M], mp[M][M];
int to[1 << M], cnt[1 << M];
int tmp[1 << M], f[N][1 << M], w[M][1 << M];
vector<int> tran[1 << M];
void Dfs1(int x, int fa) {
  f[x][0] = 1;
  for (int i = 1; i <= m; i++) f[x][1 << i - 1] = 1;
  for (int i = h[x]; i; i = g[i].next) {
    int to = g[i].to;
    if (to == fa) continue;
    Dfs1(to, x);
    memcpy(tmp, f[x], sizeof(f[x]));
    memset(f[x], 0, sizeof(f[x]));
    for (int j = 0; j < (1 << m); j++) {
      if (::to[j] < 0 || !tmp[j]) continue;
      for (int k = 0; k < tran[j].size(); k++) {
        int p = tran[j][k];
        f[x][j | p] = (f[x][j | p] + (long long)tmp[j] * f[to][p]) % mod;
      }
    }
  }
}
void Dfs2(int x, int fa) {
  w[x][0] = 1;
  for (int i = 1; i <= m; i++) w[x][1 << i - 1] = 1;
  for (int to = 1; to <= m; to++) {
    if (!mp[x][to] || to == fa) continue;
    Dfs2(to, x);
    memcpy(tmp, w[x], sizeof(w[x]));
    memset(w[x], 0, sizeof(w[x]));
    for (int j = 0; j < (1 << m); j++) {
      if (::to[j] < 0 || !tmp[j]) continue;
      for (int k = 0; k < tran[j].size(); k++) {
        int p = tran[j][k];
        w[x][j | p] = (w[x][j | p] + (long long)tmp[j] * w[to][p]) % mod;
      }
    }
  }
}
int ksm(int x, int k) {
  int ret = 1;
  while (k) {
    if (k & 1) ret = (long long)ret * x % mod;
    x = (long long)x * x % mod, k >>= 1;
  }
  return ret;
}
int main() {
  n = Getint();
  for (int i = 1; i < n; i++) {
    int x = Getint(), y = Getint();
    AddEdge(x, y), AddEdge(y, x);
  }
  m = Getint();
  for (int i = 1; i < m; i++) {
    int x = Getint(), y = Getint();
    mp[x][y] = mp[y][x] = 1;
    rd[x]++, rd[y]++;
  }
  for (int S = 0; S < (1 << m); S++) {
    int tmp[M];
    memcpy(tmp, rd, sizeof(rd));
    for (int i = 1; i <= m; i++) {
      if (!(S & (1 << i - 1))) continue;
      for (int j = i + 1; j <= m; j++)
        if (S & (1 << j - 1) && mp[i][j]) tmp[i]--, tmp[j]--;
    }
    int pos = 0;
    for (int i = 1; i <= m; i++)
      if (S & (1 << i - 1) && tmp[i] > 0) pos = pos ? -1 : i;
    to[S] = pos, cnt[S] = tmp[pos];
  }
  int S = (1 << m) - 1;
  tran[0].push_back(S);
  for (int i = 0; i < (1 << m); i++) {
    if (to[i] < 0) continue;
    tran[i].push_back(0);
    for (int j = S ^ i; j; j = (j - 1) & (S ^ i)) {
      if (to[j] < 0 || !mp[to[i]][to[j]] || to[i | j] < 0 || cnt[j] > 1)
        continue;
      tran[i].push_back(j);
    }
  }
  Dfs1(1, 0), Dfs2(1, 0);
  cout << (long long)f[1][S] * ksm(w[1][S], mod - 2) % mod;
  return 0;
}