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 f[1005][(1 << 12) + 5];
struct edge {
  int to, next;
} e[1005 * 2];
struct Tree {
  int c[1005][1005], head[1005];
  int sz[1005], son[1005], hsh[1005];
  int n, tot;
  edge e[1005 * 2];
  void add(int x, int y) {
    e[++tot] = (edge){y, head[x]};
    head[x] = tot;
  }
  void init() {
    scanf("%d", &n);
    for (int i = 2; i <= n; i++) {
      int x, y;
      scanf("%d%d", &x, &y);
      add(x, y);
      add(y, x);
    }
  }
  void dfs(int x, int fa) {
    son[x] = hsh[x] = 0;
    for (int i = head[x]; i; i = e[i].next)
      if (e[i].to != fa) {
        c[x][son[x]++] = e[i].to;
        dfs(e[i].to, x);
        hsh[x] |= 1 << (e[i].to - 1);
      }
  }
} a, b;
int dfs(int x, int y, int v) {
  if (!y) return !v;
  int &ans = f[a.c[x][y - 1]][v];
  if (ans) return ans - 1;
  ans = dfs(x, y - 1, v);
  for (int to = a.c[x][y - 1], k = 0; k < b.n; k++)
    if ((v >> k) & 1)
      ans = (ans + 1ll * dfs(x, y - 1, v ^ (1 << k)) *
                       dfs(to, a.son[to], b.hsh[k + 1])) %
            1000000007;
  return ans++;
}
int Inv(int x) {
  return x <= 1 ? x
                : 1ll * (1000000007 - 1000000007 / x) * Inv(1000000007 % x) %
                      1000000007;
}
int main() {
  a.init();
  b.init();
  a.dfs(1, 0);
  int ans = 0, Ans = 0;
  for (int i = 1; i <= b.n; i++) {
    b.dfs(i, 0);
    for (int j = 1; j <= a.n; j++)
      ans = (ans + dfs(j, a.son[j], b.hsh[i])) % 1000000007;
    memset(f, 0, sizeof(f));
  }
  a = b;
  a.dfs(1, 0);
  for (int i = 1; i <= b.n; i++) {
    b.dfs(i, 0);
    for (int j = 1; j <= a.n; j++)
      Ans = (Ans + dfs(j, a.son[j], b.hsh[i])) % 1000000007;
    memset(f, 0, sizeof(f));
  }
  printf("%d", ans * 1ll * Inv(Ans) % 1000000007);
}