#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7, N = 1013, M = 13, O = 4097;
int inv(int x) { return x < 2 ? 1 : (P - P / x) * 1ll * inv(P % x) % P; }
int n, m, p, g[M][M], f[N][O], s, t;
vector<int> v[N];
void dfs(int x, int y) {
  g[x][y] = 1 << x - 1;
  for (int j : v[x + n])
    if (j - n != y) dfs(j - n, x), g[x][y] |= g[j - n][x];
}
void dp(int x, int y) {
  int i, k;
  for (i = 0; i < m; ++i) f[x][1 << i] = 1;
  for (int j : v[x])
    if (j != y) {
      dp(j, x);
      for (k = p - 1; ~k; --k)
        for (i = 1; i <= m; ++i)
          if (k & (1 << i - 1))
            for (int l : v[i + n])
              if (!(k & g[l - n][i]))
                f[x][k | g[l - n][i]] = (f[x][k | g[l - n][i]] +
                                         f[x][k] * 1ll * f[j][g[l - n][i]]) %
                                        P;
    }
  s = (s + f[x][p - 1]) % P;
}
int main() {
  int i, j, k;
  scanf("%d", &n);
  for (i = 1; i < n; ++i)
    scanf("%d%d", &j, &k), v[j].push_back(k), v[k].push_back(j);
  scanf("%d", &m), p = 1 << m;
  for (i = 1; i < m; ++i)
    scanf("%d%d", &j, &k), v[j + n].push_back(k + n), v[k + n].push_back(j + n);
  for (i = 1; i <= m; ++i) dfs(i, 0);
  dp(1, 0), t = s, s = 0, dp(n + 1, 0), printf("%lld", t * 1ll * inv(s) % P);
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int P = 1e9 + 7, N = 1013, M = 13, O = 4097;
int inv(int x) { return x < 2 ? 1 : (P - P / x) * 1ll * inv(P % x) % P; }
int n, m, p, g[M][M], f[N][O], s, t;
vector<int> v[N];
void dfs(int x, int y) {
  g[x][y] = 1 << x - 1;
  for (int j : v[x + n])
    if (j - n != y) dfs(j - n, x), g[x][y] |= g[j - n][x];
}
void dp(int x, int y) {
  int i, k;
  for (i = 0; i < m; ++i) f[x][1 << i] = 1;
  for (int j : v[x])
    if (j != y) {
      dp(j, x);
      for (k = p - 1; ~k; --k)
        for (i = 1; i <= m; ++i)
          if (k & (1 << i - 1))
            for (int l : v[i + n])
              if (!(k & g[l - n][i]))
                f[x][k | g[l - n][i]] = (f[x][k | g[l - n][i]] +
                                         f[x][k] * 1ll * f[j][g[l - n][i]]) %
                                        P;
    }
  s = (s + f[x][p - 1]) % P;
}
int main() {
  int i, j, k;
  scanf("%d", &n);
  for (i = 1; i < n; ++i)
    scanf("%d%d", &j, &k), v[j].push_back(k), v[k].push_back(j);
  scanf("%d", &m), p = 1 << m;
  for (i = 1; i < m; ++i)
    scanf("%d%d", &j, &k), v[j + n].push_back(k + n), v[k + n].push_back(j + n);
  for (i = 1; i <= m; ++i) dfs(i, 0);
  dp(1, 0), t = s, s = 0, dp(n + 1, 0), printf("%lld", t * 1ll * inv(s) % P);
  return 0;
}

```