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 INF = 0x3f3f3f3f, N = 2e6 + 5, MOD = 1e9 + 7;
long long qpow(long long a, long long b) {
  long long res = 1;
  for (; b; b >>= 1, a = a * a % MOD)
    if (b & 1) res = res * a % MOD;
  return res;
}
class Tree {
 public:
  int n, siz[N], bits[N];
  vector<int> g[N], st[N];
  void read() {
    cin >> n;
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v), g[v].push_back(u);
    }
  }
  void dfs(int x, int _fa) {
    st[x].clear();
    st[x].push_back(0);
    siz[x] = 0;
    bits[x] = 0;
    for (auto v : g[x])
      if (v != _fa) {
        dfs(v, x);
        st[x].push_back(v);
        siz[x]++;
        bits[x] |= 1 << (v - 1);
      }
  }
} s, t;
int vis[1005][(1 << 12) + 5], dp[1005][(1 << 12) + 5];
int dfs(int x, int th, int sta) {
  if (!th) return !sta;
  int son = s.st[x][th], &res = dp[son][sta];
  if (vis[son][sta]) return dp[son][sta];
  vis[son][sta] = 1;
  res = dfs(x, th - 1, sta);
  for (int i = 1; i <= t.n; i++)
    if (sta & (1 << (i - 1))) {
      res = (res + 1ll * dfs(x, th - 1, sta ^ (1 << (i - 1))) *
                       dfs(son, s.siz[son], t.bits[i])) %
            MOD;
    }
  return res;
}
int Solve(int cas = 0) {
  s.read();
  t.read();
  s.dfs(1, 0);
  long long ans = 0, ans2 = 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 = (0ll + ans + dfs(j, s.siz[j], t.bits[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.siz[1], t.bits[i])) % MOD;
  }
  cout << ans * qpow(ans2, MOD - 2) % MOD << endl;
  return 0;
}
void Pre() {}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  Pre();
  while (Solve())
    ;
  return 0;
}