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;
const int N = 2005, mod = 1e9 + 7;
int n, m, tot, h[N], f[N][1 << 12], in[15], to[1 << 12], ct[1 << 12],
    tmp[1 << 12];
bool mp[15][15];
vector<int> tran[1 << 12];
struct edge {
  int to, next;
} e[N * 2];
void addedge(int x, int y) {
  static int cnt;
  e[++cnt] = (edge){y, h[x]};
  h[x] = cnt;
}
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 y = 1; y <= m; y++) {
    if (!mp[x][y] || y == fa) continue;
    dfs1(y, 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[y][p]) % mod;
      }
    }
  }
}
void dfs2(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 = e[i].next) {
    int y = e[i].to;
    if (y == fa) continue;
    dfs2(y, 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[y][p]) % mod;
      }
    }
  }
}
int Pow(int x, int k) {
  int ret = 1;
  while (k) {
    if (k & 1) ret = (long long)ret * x % mod;
    k >>= 1;
    x = (long long)x * x % mod;
  }
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    addedge(x, y);
    addedge(y, x);
  }
  scanf("%d", &m);
  for (int i = 1; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    mp[x][y] = mp[y][x] = 1;
    in[x]++;
    in[y]++;
  }
  for (int S = 0; S < (1 << m); S++) {
    int tmp[15];
    memcpy(tmp, in, sizeof(in));
    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) {
        if (!pos)
          pos = i;
        else
          pos = -1;
      }
    to[S] = pos;
    ct[S] = tmp[pos];
  }
  tran[0].push_back((1 << m) - 1);
  int S = (1 << m) - 1;
  for (int i = 0; i < (1 << m); i++) {
    tran[i].push_back(0);
    if (to[i] < 0) continue;
    for (int j = S ^ i; j; j = (j - 1) & (S ^ i))
      if (to[j] >= 0 && mp[to[i]][to[j]] && to[i | j] >= 0 && ct[j] <= 1) {
        tran[i].push_back(j);
      }
  }
  dfs1(1, 0);
  tot = f[1][(1 << m) - 1];
  memset(f, 0, sizeof(f));
  dfs2(1, 0);
  cout << (long long)f[1][(1 << m) - 1] * Pow(tot, mod - 2) % mod << "\n";
}