#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1005, P = 1000000007;
ll qpow(ll a, ll b) {
  ll ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % P;
    a = a * a % P;
    b >>= 1;
  }
  return ret;
}
inline ll inv(int a) { return qpow(a, P - 2); }
struct Tree {
  vector<int> v[N];
  int n, sz[N], msk[N], son[N][N];
  void read() {
    cin >> n;
    int t1, t2;
    for (int i = 1; i < n; i++)
      cin >> t1 >> t2, v[t1].push_back(t2), v[t2].push_back(t1);
  }
  void dfs(int pos, int fa) {
    sz[pos] = msk[pos] = 0;
    for (auto &i : v[pos])
      if (i != fa) dfs(son[pos][++sz[pos]] = i, pos), msk[pos] |= 1 << (i - 1);
  }
} S, T;
ll dp[N][4096], ans1, ans2;
bool vis[N][4096];
ll dfs(int pos, int msk, int cnt) {
  if (!cnt) return !msk;
  int cur = S.son[pos][cnt];
  if (vis[cur][msk]) return dp[cur][msk];
  vis[cur][msk] = 1;
  ll &ret = dp[cur][msk];
  ret = dfs(pos, msk, cnt - 1);
  for (int i = 0; i < T.n; i++)
    if (msk & (1 << i))
      ret = (ret + dfs(pos, msk ^ (1 << i), cnt - 1) *
                       dfs(cur, T.msk[i + 1], S.sz[cur])) %
            P;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  S.read();
  T.read();
  S.dfs(1, 0);
  for (int i = 1; i <= T.n; i++) {
    memset(dp, 0, sizeof(dp));
    memset(vis, 0, sizeof(vis));
    T.dfs(i, 0);
    for (int j = 1; j <= S.n; j++)
      ans1 = (ans1 + dfs(j, T.msk[i], S.sz[j])) % P;
  }
  S = T;
  S.dfs(1, 0);
  for (int i = 1; i <= T.n; i++) {
    memset(dp, 0, sizeof(dp));
    memset(vis, 0, sizeof(vis));
    T.dfs(i, 0);
    ans2 = (ans2 + dfs(1, T.msk[i], S.sz[1])) % P;
  }
  cout << ans1 * inv(ans2) % P << endl;
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1005, P = 1000000007;
ll qpow(ll a, ll b) {
  ll ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % P;
    a = a * a % P;
    b >>= 1;
  }
  return ret;
}
inline ll inv(int a) { return qpow(a, P - 2); }
struct Tree {
  vector<int> v[N];
  int n, sz[N], msk[N], son[N][N];
  void read() {
    cin >> n;
    int t1, t2;
    for (int i = 1; i < n; i++)
      cin >> t1 >> t2, v[t1].push_back(t2), v[t2].push_back(t1);
  }
  void dfs(int pos, int fa) {
    sz[pos] = msk[pos] = 0;
    for (auto &i : v[pos])
      if (i != fa) dfs(son[pos][++sz[pos]] = i, pos), msk[pos] |= 1 << (i - 1);
  }
} S, T;
ll dp[N][4096], ans1, ans2;
bool vis[N][4096];
ll dfs(int pos, int msk, int cnt) {
  if (!cnt) return !msk;
  int cur = S.son[pos][cnt];
  if (vis[cur][msk]) return dp[cur][msk];
  vis[cur][msk] = 1;
  ll &ret = dp[cur][msk];
  ret = dfs(pos, msk, cnt - 1);
  for (int i = 0; i < T.n; i++)
    if (msk & (1 << i))
      ret = (ret + dfs(pos, msk ^ (1 << i), cnt - 1) *
                       dfs(cur, T.msk[i + 1], S.sz[cur])) %
            P;
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  S.read();
  T.read();
  S.dfs(1, 0);
  for (int i = 1; i <= T.n; i++) {
    memset(dp, 0, sizeof(dp));
    memset(vis, 0, sizeof(vis));
    T.dfs(i, 0);
    for (int j = 1; j <= S.n; j++)
      ans1 = (ans1 + dfs(j, T.msk[i], S.sz[j])) % P;
  }
  S = T;
  S.dfs(1, 0);
  for (int i = 1; i <= T.n; i++) {
    memset(dp, 0, sizeof(dp));
    memset(vis, 0, sizeof(vis));
    T.dfs(i, 0);
    ans2 = (ans2 + dfs(1, T.msk[i], S.sz[1])) % P;
  }
  cout << ans1 * inv(ans2) % P << endl;
  return 0;
}

```