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 MOD = 1e9 + 7;
long long Pow(long long x, long long y) {
  long long s = 1;
  for (; y; y >>= 1, x = (long long)x * x % MOD)
    if (y & 1) s = (long long)s * x % MOD;
  return s;
}
struct edge {
  int to, nxt;
};
struct tree {
  edge e[2010];
  int n, cnt;
  int head[1010], son[1010][1010], size[1010], bit[1010];
  void addedge(int x, int y) {
    e[++cnt].to = y;
    e[cnt].nxt = head[x];
    head[x] = cnt;
  }
  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);
    }
  }
  void dfs(int x, int fa) {
    size[x] = bit[x] = 0;
    for (int i = head[x]; i; i = e[i].nxt)
      if (e[i].to ^ fa)
        dfs(son[x][++size[x]] = e[i].to, x), bit[x] |= 1 << e[i].to - 1;
  }
} S, T;
int dp[1010][1 << 12], ans, ans2;
bool vis[1010][1 << 12];
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 = ((long long)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 += 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 += dfs(1, S.size[1], T.bit[i])) %= MOD;
  }
  ans = (long long)ans * Pow(ans2, MOD - 2) % MOD;
  printf("%d", ans);
  getchar();
  getchar();
  return 0;
}