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 mod = 1e9 + 7;
int n, m, cnt1, cnt2, ans, ans2;
int size1[2000 + 10], size2[2000 + 10], size3[2000 + 10];
int to1[2000 * 3 + 10], nxt1[2000 * 3 + 10], head1[2000 + 10];
int to2[2000 * 3 + 10], nxt2[2000 * 3 + 10], head2[2000 + 10];
int f[2000 + 10][(1 << 12) + 10], vis[2000 + 10][(1 << 12) + 10];
int state1[2000 + 10], state2[2000 + 10], state3[2000 + 10];
int son1[2000 + 10][2000 + 10], son2[2000 + 10][2000 + 10],
    son3[2000 + 10][2000 + 10];
void add1(int u, int v) {
  to1[++cnt1] = v;
  nxt1[cnt1] = head1[u];
  head1[u] = cnt1;
}
void add2(int u, int v) {
  to2[++cnt2] = v;
  nxt2[cnt2] = head2[u];
  head2[u] = cnt2;
}
void dfs1(int x, int ff) {
  size1[x] = state1[x] = 0;
  for (int i = head1[x]; i; i = nxt1[i]) {
    int tox = to1[i];
    if (tox == ff) continue;
    son1[x][++size1[x]] = tox;
    state1[x] |= (1 << (tox - 1));
    dfs1(tox, x);
  }
}
void dfs2(int x, int ff) {
  size2[x] = state2[x] = 0;
  for (int i = head2[x]; i; i = nxt2[i]) {
    int tox = to2[i];
    if (tox == ff) continue;
    son2[x][++size2[x]] = tox;
    state2[x] |= (1 << (tox - 1));
    dfs2(tox, x);
  }
}
int dfs(int ff, int x, int y) {
  if (!x) return !y;
  int k = son1[ff][x];
  if (vis[k][y]) return f[k][y];
  vis[k][y] = 1;
  f[k][y] = dfs(ff, x - 1, y);
  for (int i = 0; i < m; i++)
    if (y >> i & 1)
      f[k][y] = (1ll * dfs(ff, x - 1, y ^ (1 << i)) *
                     dfs(k, size1[k], state2[i + 1]) +
                 f[k][y]) %
                mod;
  return f[k][y];
}
int qpow(int x, int y) {
  int cnt = 1;
  while (y) {
    if (y & 1) cnt = 1ll * cnt * x % mod;
    x = 1ll * x * x % mod;
    y >>= 1;
  }
  return cnt;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add1(u, v);
    add1(v, u);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add2(u, v);
    add2(v, u);
  }
  dfs1(1, 0);
  for (int i = 1; i <= m; i++) {
    memset(f, 0, sizeof(f));
    memset(vis, 0, sizeof(vis));
    dfs2(i, 0);
    for (int j = 1; j <= n; j++)
      ans = (ans + dfs(j, size1[j], state2[i])) % mod;
  }
  for (int i = 1; i <= cnt1; i++) head1[i] = 0;
  cnt1 = cnt2;
  for (int i = 1; i <= cnt2; i++) {
    to1[i] = to2[i];
    nxt1[i] = nxt2[i];
  }
  for (int i = 1; i <= m; i++) head1[i] = head2[i];
  dfs1(1, 0);
  for (int i = 1; i <= m; i++) {
    memset(f, 0, sizeof(f));
    memset(vis, 0, sizeof(vis));
    dfs2(i, 0);
    ans2 = (ans2 + dfs(1, size1[1], state2[i])) % mod;
  }
  ans = 1ll * ans * qpow(ans2, mod - 2) % mod;
  printf("%d\n", ans);
  return 0;
}