#include <bits/stdc++.h>
using namespace std;
const int N = 1010, M = (1 << 12), MOD = 1e9 + 7;
int n, m, ans, f[N][M];
struct edge {
  int next, to;
};
long long fpow(long long x, long long k) {
  long long ans = 1;
  for (; k; k >>= 1, x = x * x % MOD)
    if (k & 1) ans = ans * x % MOD;
  return ans;
}
struct Tree1 {
  int tot, head[N], bit[N];
  edge e[N * 2];
  void add(int from, int to) {
    e[++tot] = (edge){head[from], to};
    head[from] = tot;
  }
  void dfs(int x, int fa) {
    bit[x] = 0;
    for (int i = head[x]; ~i; i = e[i].next) {
      int v = e[i].to;
      if (v != fa) dfs(v, x), bit[x] |= (1 << v - 1);
    }
  }
} T;
struct Tree2 {
  int tot, head[N], pre[N], son[N];
  edge e[N * 2];
  void add(int from, int to) {
    e[++tot] = (edge){head[from], to};
    head[from] = tot;
  }
  void dfs1(int x, int fa) {
    int last = 0;
    son[x] = 0;
    for (int i = head[x]; ~i; i = e[i].next) {
      int v = e[i].to;
      if (v != fa) {
        pre[v] = last;
        son[x] = v;
        last = v;
        dfs1(v, x);
      }
    }
  }
  int dfs2(int x, int s) {
    if (!s) return 1;
    if (!x) return 0;
    if (f[x][s] != -1) return f[x][s];
    f[x][s] = dfs2(pre[x], s);
    for (int i = 1; i <= m; i++)
      if (s & (1 << i - 1))
        f[x][s] = (f[x][s] + 1LL * dfs2(son[x], T.bit[i]) *
                                 dfs2(pre[x], s ^ (1 << i - 1))) %
                  MOD;
    return f[x][s];
  }
  int solve() {
    int ans = 0;
    dfs1(1, 0);
    for (int i = 1; i <= m; i++) {
      memset(f, -1, sizeof(f));
      T.dfs(i, 0);
      for (int j = 1; j <= n; j++) ans = (ans + dfs2(son[j], T.bit[i])) % MOD;
    }
    return ans;
  }
} S;
int main() {
  scanf("%d", &n);
  memset(S.head, -1, sizeof(S.head));
  memset(T.head, -1, sizeof(T.head));
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    S.add(x, y);
    S.add(y, x);
  }
  scanf("%d", &m);
  for (int i = 1, x, y; i < m; i++) {
    scanf("%d%d", &x, &y);
    T.add(x, y);
    T.add(y, x);
  }
  ans = S.solve();
  n = m;
  memcpy(S.e, T.e, sizeof(S.e));
  memcpy(S.head, T.head, sizeof(S.head));
  printf("%lld", ans * fpow(S.solve(), MOD - 2) % MOD);
  return 0;
}

### Prompt

Develop a solution in Cpp 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 N = 1010, M = (1 << 12), MOD = 1e9 + 7;
int n, m, ans, f[N][M];
struct edge {
  int next, to;
};
long long fpow(long long x, long long k) {
  long long ans = 1;
  for (; k; k >>= 1, x = x * x % MOD)
    if (k & 1) ans = ans * x % MOD;
  return ans;
}
struct Tree1 {
  int tot, head[N], bit[N];
  edge e[N * 2];
  void add(int from, int to) {
    e[++tot] = (edge){head[from], to};
    head[from] = tot;
  }
  void dfs(int x, int fa) {
    bit[x] = 0;
    for (int i = head[x]; ~i; i = e[i].next) {
      int v = e[i].to;
      if (v != fa) dfs(v, x), bit[x] |= (1 << v - 1);
    }
  }
} T;
struct Tree2 {
  int tot, head[N], pre[N], son[N];
  edge e[N * 2];
  void add(int from, int to) {
    e[++tot] = (edge){head[from], to};
    head[from] = tot;
  }
  void dfs1(int x, int fa) {
    int last = 0;
    son[x] = 0;
    for (int i = head[x]; ~i; i = e[i].next) {
      int v = e[i].to;
      if (v != fa) {
        pre[v] = last;
        son[x] = v;
        last = v;
        dfs1(v, x);
      }
    }
  }
  int dfs2(int x, int s) {
    if (!s) return 1;
    if (!x) return 0;
    if (f[x][s] != -1) return f[x][s];
    f[x][s] = dfs2(pre[x], s);
    for (int i = 1; i <= m; i++)
      if (s & (1 << i - 1))
        f[x][s] = (f[x][s] + 1LL * dfs2(son[x], T.bit[i]) *
                                 dfs2(pre[x], s ^ (1 << i - 1))) %
                  MOD;
    return f[x][s];
  }
  int solve() {
    int ans = 0;
    dfs1(1, 0);
    for (int i = 1; i <= m; i++) {
      memset(f, -1, sizeof(f));
      T.dfs(i, 0);
      for (int j = 1; j <= n; j++) ans = (ans + dfs2(son[j], T.bit[i])) % MOD;
    }
    return ans;
  }
} S;
int main() {
  scanf("%d", &n);
  memset(S.head, -1, sizeof(S.head));
  memset(T.head, -1, sizeof(T.head));
  for (int i = 1, x, y; i < n; i++) {
    scanf("%d%d", &x, &y);
    S.add(x, y);
    S.add(y, x);
  }
  scanf("%d", &m);
  for (int i = 1, x, y; i < m; i++) {
    scanf("%d%d", &x, &y);
    T.add(x, y);
    T.add(y, x);
  }
  ans = S.solve();
  n = m;
  memcpy(S.e, T.e, sizeof(S.e));
  memcpy(S.head, T.head, sizeof(S.head));
  printf("%lld", ans * fpow(S.solve(), MOD - 2) % MOD);
  return 0;
}

```