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 = 13, M = 1e3 + 10, P = 1e9 + 7;
vector<int> dp[M][N];
int Log2[1 << 12 | 5];
void fix(int& x) {
  if (x >= P) x -= P;
  if (x < 0) x += P;
}
int power(int a, int b) {
  int c = 1;
  for (; b; b >>= 1, a = 1ll * a * a % P)
    if (b & 1) c = 1ll * a * c % P;
  return c;
}
void rdtr(vector<vector<int> >& x) {
  int siz;
  cin >> siz;
  vector<vector<int> >().swap(x);
  x.resize(siz);
  for (int i = 1, u, v; i < siz; i++) {
    scanf("%d%d", &u, &v);
    u--, v--;
    x[u].push_back(v), x[v].push_back(u);
  }
}
void printtr(vector<vector<int> > x, int rt) {
  queue<int> q;
  q.push(rt);
  while (q.size()) {
    int nx = q.front();
    q.pop();
    for (auto ny : x[nx]) {
      printf("%d - > %d\n", nx, ny);
      q.push(ny);
    }
  }
}
void mkrt(vector<vector<int> >& x, int rt) {
  queue<int> q;
  q.push(rt);
  while (q.size()) {
    int nx = q.front();
    q.pop();
    for (int i = x[nx].size(), ny; i; i--) {
      ny = x[nx][i - 1];
      for (vector<int>::iterator f = x[ny].begin(); f != x[ny].end(); ++f)
        if ((*f) == nx) {
          x[ny].erase(f);
          break;
        }
      q.push(ny);
    }
  }
}
int DP(vector<vector<int> >& S, vector<vector<int> >& T, int x, int y) {
  if (dp[x][y].size()) return dp[x][y].back();
  int siz = 1 << T[y].size();
  dp[x][y].resize(siz);
  dp[x][y][0] = 1;
  for (int u : S[x])
    for (int j = siz - 1; j >= 0; j--) {
      for (int k = j; k; k ^= k & -k) {
        int v = Log2[k & -k];
        if (dp[x][y][j ^ (1 << v)])
          fix(dp[x][y][j] +=
              1ll * dp[x][y][j ^ (1 << v)] * DP(S, T, u, T[y][v]) % P);
      }
    }
  return dp[x][y].back();
}
int match(vector<vector<int> > s, vector<vector<int> > t) {
  int ans = 0;
  mkrt(s, 0);
  for (int i = 0; i < (int)t.size(); i++) {
    vector<vector<int> > x = t;
    mkrt(x, i);
    for (int a = 0; a < (int)s.size(); a++)
      for (int b = 0; b < (int)t.size(); b++) vector<int>().swap(dp[a][b]);
    for (int j = 0; j < (int)s.size(); j++) {
      fix(ans += DP(s, x, j, i));
    }
  }
  return ans;
}
signed main() {
  vector<vector<int> > s, t;
  rdtr(s), rdtr(t);
  for (int i = 1, k = 0; k <= 12; k++, i <<= 1) Log2[i] = k;
  int ans = match(s, t), tmp = match(t, t);
  ans = 1ll * ans * power(tmp, P - 2) % P;
  cout << ans << endl;
  return 0;
}