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;
int len(int x) { return 32 - __builtin_clz(x); }
int Union(int x, int y) { return (x << len(y)) | y; }
struct Tree {
  int n, ans;
  int f[1 << 12][2];
  vector<int> d[1001];
  map<int, int> num[1001];
  map<int, vector<int> > mp;
  void read() {
    scanf("%d", &n);
    for (int i = 2; i <= n; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      d[u].push_back(v);
      d[v].push_back(u);
    }
  }
  int dfs(int cur, int pre) {
    int res = 1;
    vector<int> tmp;
    for (auto nxt : d[cur])
      if (nxt != pre) tmp.push_back(dfs(nxt, cur));
    sort(tmp.begin(), tmp.end());
    for (auto x : tmp) res = Union(res, x);
    res <<= 1;
    if (!mp.count(res)) mp[res] = tmp;
    return res;
  }
  void getID() {
    for (int i = 1; i <= n; i++) dfs(i, 0);
  }
  void DP(int cur, int pre, const Tree &T) {
    for (auto nxt : d[cur])
      if (nxt != pre) DP(nxt, cur, T);
    for (const auto &pi : T.mp) {
      auto &types = pi.second;
      int id = pi.first, n = types.size(), now = 0, lst = 1;
      for (int i = 0; i < (1 << n); i++) f[i][0] = f[i][1] = 0;
      f[0][0] = 1;
      for (auto nxt : d[cur])
        if (nxt != pre) {
          now ^= 1, lst ^= 1;
          for (int i = 0; i < (1 << n); i++) f[i][now] = f[i][lst];
          for (int i = 0; i < n; i++)
            if (num[nxt][types[i]])
              for (int j = (1 << n) - 1; j >= 0; j--)
                if (f[j][lst] && !((1 << i) & j) &&
                    !(i && types[i] == types[i - 1] && !((1 << (i - 1)) & j)))
                  (f[(1 << i) | j][now] +=
                   1ll * f[j][lst] * num[nxt][types[i]] % 1000000007) %=
                      1000000007;
        }
      if (f[(1 << n) - 1][now]) {
        num[cur][id] = f[(1 << n) - 1][now];
        if (len(id) == 2 * T.n) (ans += num[cur][id]) %= 1000000007;
      }
    }
  }
} S, T;
int main() {
  S.read();
  T.read();
  T.getID();
  S.DP(1, 0, T);
  printf("%d\n", S.ans);
}