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 N = 1005;
const long long mo = 1e9 + 7;
struct edg {
  int to, nxt;
};
struct tree {
  edg e[N << 1];
  int siz[N], bit[N], son[N][N], head[N], n, cnt;
  inline void add(int u, int v) {
    cnt++;
    e[cnt].to = v;
    e[cnt].nxt = head[u];
    head[u] = cnt;
  }
  inline void read() {
    scanf("%d", &n);
    for (int i = 1; i < n; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      add(v, u);
      add(u, v);
    }
  }
  void dfs(int u, int f) {
    siz[u] = bit[u] = 0;
    for (int i = head[u]; i; i = e[i].nxt) {
      int v = e[i].to;
      if (v == f) continue;
      son[u][++siz[u]] = v;
      dfs(v, u);
      bit[u] |= 1 << (v - 1);
    }
  }
} S, T;
long long getinv(int x) {
  return x <= 1 ? 1 : (long long)(mo - mo / x) * getinv(mo % x) % mo;
}
const int M = (1 << 12) + 5;
long long dp[N][M], ans1, ans2;
bool vis[N][M];
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;
  long long &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.siz[x], T.bit[i + 1]) +
           ret) %
          mo;
  return ret;
}
int main() {
  S.read(), T.read();
  S.dfs(1, 0);
  for (int i = 1; i <= T.n; i++) {
    memset(vis, 0, sizeof(vis));
    memset(dp, 0, sizeof(dp));
    T.dfs(i, 0);
    for (int j = 1; j <= S.n; j++)
      ans1 = (ans1 + dfs(j, S.siz[j], T.bit[i])) % mo;
  }
  S = T;
  S.dfs(1, 0);
  for (int i = 1; i <= T.n; i++) {
    memset(vis, 0, sizeof(vis));
    memset(dp, 0, sizeof(dp));
    T.dfs(i, 0);
    ans2 = (ans2 + dfs(1, S.siz[1], T.bit[i])) % mo;
  }
  long long ans = ans1 * getinv(ans2) % mo;
  printf("%lld", ans);
  return 0;
}