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 mod = 1e9 + 7;
struct edge {
  int v, next;
};
int ans, ans2, f[1005][1 << 12];
int Pow(int a, int b) {
  int rec = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) rec = 1ll * rec * a % mod;
  return rec;
}
struct nade {
  int head[1005], edgenum, d[1005][1005], num[1005], sta[1005], n;
  edge e[1005 << 1];
  void add(int u, int v) {
    e[++edgenum] = edge{v, head[u]};
    head[u] = edgenum;
  }
  void dfs(int u, int fa) {
    num[u] = sta[u] = 0;
    for (int i = head[u], v; i; i = e[i].next)
      if ((v = e[i].v) ^ fa) {
        dfs(d[u][num[u]++] = v, u);
        sta[u] |= 1 << v - 1;
      }
  }
} s, t;
int dfs(int u, int i, int sta) {
  if (!i) return !sta;
  int &rec = f[s.d[u][i - 1]][sta];
  if (rec != -1) return rec;
  rec = dfs(u, i - 1, sta);
  for (int v = s.d[u][i - 1], k = 0; k < t.n; k++)
    if (sta >> k & 1)
      rec = (rec + 1ll * dfs(u, i - 1, sta ^ (1 << k)) *
                       dfs(v, s.num[v], t.sta[k + 1])) %
            mod;
  return rec;
}
int main() {
  scanf("%d", &s.n);
  for (int i = 1, u, v; i < s.n; i++) {
    scanf("%d%d", &u, &v);
    s.add(u, v);
    s.add(v, u);
  }
  scanf("%d", &t.n);
  for (int i = 1, u, v; i < t.n; i++) {
    scanf("%d%d", &u, &v);
    t.add(u, v);
    t.add(v, u);
  }
  s.dfs(1, 0);
  memset(f, -1, sizeof f);
  for (int r = 1; r <= t.n; r++) {
    t.dfs(r, 0);
    for (int i = 1; i <= s.n; i++)
      ans = (ans + dfs(i, s.num[i], t.sta[r])) % mod;
    memset(f, -1, sizeof f);
  }
  s = t;
  s.dfs(1, 0);
  for (int r = 1; r <= t.n; r++) {
    t.dfs(r, 0);
    ans2 = (ans2 + dfs(1, s.num[1], t.sta[r])) % mod;
    memset(f, -1, sizeof f);
  }
  ans = 1ll * ans * Pow(ans2, mod - 2) % mod;
  printf("%d", ans);
}