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 MaxN = 1001;
const int MaxV = 1 << 12;
const int MOD = 1e9 + 7;
inline int Pow(int x, int y) {
  int s = 1;
  for (; y; y >>= 1, x = 1ll * x * x % MOD) {
    if (y & 1) {
      s = 1ll * s * x % MOD;
    }
  }
  return s;
}
inline int inv(int x) { return Pow(x, MOD - 2); }
struct Edge {
  int x, y;
  void set(int x, int y) {
    this->x = x;
    this->y = y;
  }
};
struct Tree {
  Edge edge[MaxN << 1];
  int n, cnt;
  int head[MaxN], son[MaxN][MaxN], size[MaxN], bit[MaxN];
  inline void addEdge(int x, int y) {
    edge[++cnt].set(head[x], y);
    head[x] = cnt;
  }
  inline void read() {
    scanf("%d", &n);
    for (int i = 1; i < n; ++i) {
      int x, y;
      scanf("%d %d", &x, &y);
      addEdge(x, y);
      addEdge(y, x);
    }
  }
  inline void dfs(int x, int fa) {
    size[x] = bit[x] = 0;
    for (int i = head[x]; i; i = edge[i].x)
      if (edge[i].y ^ fa) {
        son[x][++size[x]] = edge[i].y;
        dfs(edge[i].y, x);
        bit[x] |= 1 << (edge[i].y - 1);
      }
  }
} S, T;
int dp[MaxN][MaxV], ans, ans2;
bool vis[MaxN][MaxV];
inline 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.size[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 = (ans + dfs(j, S.size[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 = (ans2 + dfs(1, S.size[1], T.bit[i])) % MOD;
  }
  ans = 1ll * ans * inv(ans2) % MOD;
  printf("%d\n", ans);
  return 0;
}