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 N = 1e3 + 10, K = 1 << 12;
const int mod = 1e9 + 7;
int s, t, dp[N][K], ans, self;
bool vis[N][K];
inline int read() {
  int s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
struct mdzz {
  int n, fst[N], nxt[N << 1], to[N << 1], tot;
  int siz[N], son[N], sons[N][N];
  inline void readd() {
    n = read();
    for (int i = 1; i < n; ++i) {
      int u = read(), v = read();
      add(u, v);
      add(v, u);
    }
  }
  inline void add(int u, int v) {
    nxt[++tot] = fst[u];
    to[tot] = v;
    fst[u] = tot;
  }
  inline void dfs(int u, int fa) {
    siz[u] = son[u] = 0;
    for (int i = fst[u]; i; i = nxt[i]) {
      int v = to[i];
      if (v == fa) continue;
      sons[u][++siz[u]] = v;
      dfs(v, u);
      son[u] |= 1 << (v - 1);
    }
  }
} S, T;
inline int ksm(int a, int b) {
  int tmp = 1;
  while (b) {
    if (b & 1) tmp = (1ll * tmp * a) % mod;
    a = (1ll * a * a) % mod;
    b >>= 1;
  }
  return tmp;
}
inline int dfss(int u, int num, int V) {
  if (num == 0) return !V;
  int v = S.sons[u][num];
  if (vis[v][V]) return dp[v][V];
  vis[v][V] = 1;
  dp[v][V] = dfss(u, num - 1, V);
  for (int i = 0; i < T.n; ++i) {
    if (V >> i & 1) {
      int ret1 = dfss(u, num - 1, V ^ (1 << i));
      int ret2 = dfss(v, S.siz[v], T.son[i + 1]);
      dp[v][V] = (dp[v][V] + 1ll * ret1 * ret2) % mod;
    }
  }
  return dp[v][V];
}
int main() {
  S.readd();
  T.readd();
  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 = (ans + dfss(j, S.siz[j], T.son[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);
    self = (self + dfss(1, S.siz[1], T.son[i])) % mod;
  }
  ans = 1ll * ans * ksm(self, mod - 2) % mod;
  printf("%d\n", ans);
  return 0;
}