#include <bits/stdc++.h>
using namespace std;
const int maxn = 1010;
const int maxv = 1 << 12;
const int mod = 1e9 + 7;
int qsm(int a, int n) {
  int ret = 1;
  while (n) {
    if (n & 1) ret = 1LL * ret * a % mod;
    a = 1LL * a * a % mod;
    n >>= 1;
  }
  return ret;
}
struct edge {
  int to, next;
};
struct tree {
  edge e[maxn << 1];
  int n, edgecnt;
  int head[maxn], son[maxn][maxn], sz[maxn], bit[maxn];
  void init() {
    memset(head, 0, sizeof(head));
    edgecnt = 0;
  }
  void add(int x, int y) {
    e[++edgecnt].to = y, e[edgecnt].next = head[x], head[x] = edgecnt;
  }
  void read() {
    init();
    scanf("%d", &n);
    for (int i = 1; i < n; i++) {
      int x, y;
      scanf("%d %d", &x, &y);
      add(x, y);
      add(y, x);
    }
  }
  void dfs(int x, int fa) {
    sz[x] = bit[x] = 0;
    for (int i = head[x]; i; i = e[i].next) {
      int v = e[i].to;
      if (v == fa) continue;
      dfs(son[x][++sz[x]] = e[i].to, x), bit[x] |= 1 << e[i].to - 1;
    }
  }
} S, T;
int dp[maxn][maxv], ans, ans2;
bool vis[maxn][maxv];
int dfs(int f, int y, int V) {
  if (!y) return !V;
  int x = S.son[f][y];
  if (vis[x][V]) return dp[x][V];
  vis[x][V] = 1;
  int &ret = dp[x][V];
  ret = dfs(f, y - 1, V);
  for (int i = 0; i < T.n; i++)
    if (V >> i & 1)
      ret = (1LL * dfs(f, y - 1, V ^ (1 << i)) * dfs(x, S.sz[x], T.bit[i + 1]) +
             ret) %
            mod;
  return ret;
}
int main() {
  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++) (ans += dfs(j, S.sz[j], T.bit[i])) %= mod;
  }
  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 += dfs(1, S.sz[1], T.bit[i])) %= mod;
  }
  ans = (1LL * ans % mod * qsm(ans2, mod - 2) % mod) % mod;
  return 0 * printf("%d\n", ans);
}

### 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;
const int maxn = 1010;
const int maxv = 1 << 12;
const int mod = 1e9 + 7;
int qsm(int a, int n) {
  int ret = 1;
  while (n) {
    if (n & 1) ret = 1LL * ret * a % mod;
    a = 1LL * a * a % mod;
    n >>= 1;
  }
  return ret;
}
struct edge {
  int to, next;
};
struct tree {
  edge e[maxn << 1];
  int n, edgecnt;
  int head[maxn], son[maxn][maxn], sz[maxn], bit[maxn];
  void init() {
    memset(head, 0, sizeof(head));
    edgecnt = 0;
  }
  void add(int x, int y) {
    e[++edgecnt].to = y, e[edgecnt].next = head[x], head[x] = edgecnt;
  }
  void read() {
    init();
    scanf("%d", &n);
    for (int i = 1; i < n; i++) {
      int x, y;
      scanf("%d %d", &x, &y);
      add(x, y);
      add(y, x);
    }
  }
  void dfs(int x, int fa) {
    sz[x] = bit[x] = 0;
    for (int i = head[x]; i; i = e[i].next) {
      int v = e[i].to;
      if (v == fa) continue;
      dfs(son[x][++sz[x]] = e[i].to, x), bit[x] |= 1 << e[i].to - 1;
    }
  }
} S, T;
int dp[maxn][maxv], ans, ans2;
bool vis[maxn][maxv];
int dfs(int f, int y, int V) {
  if (!y) return !V;
  int x = S.son[f][y];
  if (vis[x][V]) return dp[x][V];
  vis[x][V] = 1;
  int &ret = dp[x][V];
  ret = dfs(f, y - 1, V);
  for (int i = 0; i < T.n; i++)
    if (V >> i & 1)
      ret = (1LL * dfs(f, y - 1, V ^ (1 << i)) * dfs(x, S.sz[x], T.bit[i + 1]) +
             ret) %
            mod;
  return ret;
}
int main() {
  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++) (ans += dfs(j, S.sz[j], T.bit[i])) %= mod;
  }
  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 += dfs(1, S.sz[1], T.bit[i])) %= mod;
  }
  ans = (1LL * ans % mod * qsm(ans2, mod - 2) % mod) % mod;
  return 0 * printf("%d\n", ans);
}

```