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 = 1005, MOD = 1000000007, A = 233, B = 101, C = 40961;
int n, m, x, y, a[20][20], ha[20], siz[20], tg, fa2[20], dp[N][20], f[1 << 12],
    ans;
set<int> s;
int head[N], ecnt, head2[20], ecnt2;
struct edge {
  int to, nxt;
} e[N << 1], e2[40];
inline void add(int x, int y) {
  e[++ecnt] = {y, head[x]};
  head[x] = ecnt;
}
inline void add2(int x, int y) {
  e2[++ecnt2] = {y, head2[x]};
  head2[x] = ecnt2;
}
inline int mo(int x) { return x >= MOD ? x - MOD : x; }
inline int ksm(int a, int b) {
  int c = 1;
  while (b) {
    if (b & 1) c = 1ll * c * a % MOD;
    a = 1ll * a * a % MOD;
    b >>= 1;
  }
  return c;
}
inline bool cmp(int a, int b) { return ha[a] < ha[b]; }
void dfs1(int x) {
  int y, o = 1;
  siz[x] = 1;
  for (int i = head2[x]; i; i = e2[i].nxt)
    if ((y = e2[i].to) != fa2[x]) {
      fa2[y] = x;
      dfs1(y);
      a[x][++a[x][0]] = y;
      siz[x] += siz[y];
    }
  sort(a[x] + 1, a[x] + a[x][0] + 1, cmp);
  for (register int i = 1; i <= a[x][0]; i++)
    ha[x] = ha[x] << (siz[a[x][i]] << 1) | ha[a[x][i]];
  ha[x] = ha[x] << 1 | 1;
  for (register int i = 2; i <= a[x][0]; i++) {
    if (ha[a[x][i]] != ha[a[x][i - 1]]) o = 0;
    tg = 1ll * tg * ++o % MOD;
  }
}
void dfs2(int x, int fa) {
  int y, full;
  for (int i = head[x]; i; i = e[i].nxt)
    if ((y = e[i].to) != fa) dfs2(y, x);
  for (int i = 1; i <= m; i++) {
    memset(f, 0, sizeof(f));
    f[0] = 1;
    full = (1 << a[i][0]) - 1;
    for (int j = head[x]; j; j = e[j].nxt)
      if ((y = e[j].to) != fa)
        for (register int ii = full; ii >= 0; ii--)
          for (register int k = 1; k <= a[i][0]; k++)
            if (ii & 1 << k - 1)
              f[ii] = (f[ii] + 1ll * f[ii ^ 1 << k - 1] * dp[y][a[i][k]]) % MOD;
    dp[x][i] = f[full];
  }
}
int main() {
  scanf("%d", &n);
  for (register int i = 1; i < n; i++)
    scanf("%d%d", &x, &y), add(x, y), add(y, x);
  scanf("%d", &m);
  for (register int i = 1; i < m; i++)
    scanf("%d%d", &x, &y), add2(x, y), add2(y, x);
  for (int i = 1; i <= m; i++) {
    memset(a, 0, sizeof(a));
    memset(dp, 0, sizeof(dp));
    memset(ha, 0, sizeof(ha));
    tg = 1;
    fa2[i] = i;
    dfs1(i);
    if (s.count(ha[i])) continue;
    s.insert(ha[i]);
    tg = ksm(tg, MOD - 2);
    dfs2(1, 1);
    for (register int j = 1; j <= n; j++)
      ans = (ans + 1ll * dp[j][i] * tg) % MOD;
  }
  printf("%d", ans);
}