Problem: Harry, Ron and Hermione have figured out that Helga Hufflepuff's cup is a horcrux. Through her encounter with Bellatrix Lestrange, Hermione came to know that the cup is present in Bellatrix's family vault in Gringott's Wizarding Bank. 

The Wizarding bank is in the form of a tree with total n vaults where each vault has some type, denoted by a number between 1 to m. A tree is an undirected connected graph with no cycles.

The vaults with the highest security are of type k, and all vaults of type k have the highest security.

There can be at most x vaults of highest security. 

Also, if a vault is of the highest security, its adjacent vaults are guaranteed to not be of the highest security and their type is guaranteed to be less than k.

Harry wants to consider every possibility so that he can easily find the best path to reach Bellatrix's vault. So, you have to tell him, given the tree structure of Gringotts, the number of possible ways of giving each vault a type such that the above conditions hold.

Input

The first line of input contains two space separated integers, n and m — the number of vaults and the number of different vault types possible. (1 ≤ n ≤ 105, 1 ≤ m ≤ 109).

Each of the next n - 1 lines contain two space separated integers ui and vi (1 ≤ ui, vi ≤ n) representing the i-th edge, which shows there is a path between the two vaults ui and vi. It is guaranteed that the given graph is a tree.

The last line of input contains two integers k and x (1 ≤ k ≤ m, 1 ≤ x ≤ 10), the type of the highest security vault and the maximum possible number of vaults of highest security.

Output

Output a single integer, the number of ways of giving each vault a type following the conditions modulo 109 + 7.

Examples

Input

4 2
1 2
2 3
1 4
1 2


Output

1


Input

3 3
1 2
1 3
2 1


Output

13


Input

3 1
1 2
1 3
1 1


Output

0

Note

In test case 1, we cannot have any vault of the highest security as its type is 1 implying that its adjacent vaults would have to have a vault type less than 1, which is not allowed. Thus, there is only one possible combination, in which all the vaults have type 2.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class htpe, class cmp>
using heap = priority_queue<htpe, vector<htpe>, cmp>;
template <class htpe>
using min_heap = heap<htpe, greater<htpe> >;
template <class htpe>
using max_heap = heap<htpe, less<htpe> >;
const int INF = 1791791791;
const long long INFLL = 1791791791791791791ll;
template <int input_buf_size, int output_buf_size>
class FastIO {
  char cbuf[input_buf_size + 1];
  int icur = 0;
  inline bool go_to_next_token() {
    while (cbuf[icur] == ' ' || cbuf[icur] == '\n') icur++;
    while (cbuf[icur] == 0) {
      icur = 0;
      if (fgets(cbuf, sizeof(cbuf), stdin) != cbuf) return false;
      while (cbuf[icur] == ' ' || cbuf[icur] == '\n') icur++;
    }
    return true;
  }

 public:
  string readString() {
    assert(go_to_next_token());
    string ans;
    while (cbuf[icur] != ' ' && cbuf[icur] != '\n' && cbuf[icur] != 0)
      ans.push_back(cbuf[icur++]);
    ans.shrink_to_fit();
    return ans;
  }
  template <class int_type>
  int_type readInt() {
    assert(go_to_next_token());
    int_type x = 0;
    bool m = cbuf[icur] == '-';
    if (m) icur++;
    while ('0' <= cbuf[icur] && cbuf[icur] <= '9') {
      x *= 10;
      x += (cbuf[icur] - '0');
      icur++;
    }
    if (m) x = -x;
    return x;
  }
  bool seekEof() { return !go_to_next_token(); }

 private:
  char obuf[output_buf_size + 1];
  int ocur = 0;
  inline void write_string(const char *str, size_t sz = 0) {
    if (sz == 0) sz = strlen(str);
    if (ocur + sz > output_buf_size) {
      fputs(obuf, stdout);
      fputs(str, stdout);
      ocur = 0;
      obuf[0] = 0;
      return;
    }
    strcpy(obuf + ocur, str);
    ocur += sz;
    obuf[ocur] = 0;
  }

 public:
  template <class int_type>
  void writeInt(int_type x, bool sp = true) {
    char buf[21];
    int c = 0;
    if (x < 0) {
      buf[c++] = '-';
      x = -x;
    }
    int s = c;
    if (x == 0) {
      buf[c++] = '0';
    }
    while (x > 0) {
      buf[c++] = (x % 10) + '0';
      x /= 10;
    }
    for (int i = 0; 2 * i < c - s; i++) {
      swap(buf[s + i], buf[c - 1 - i]);
    }
    buf[c] = 0;
    write_string(buf, c);
    if (sp) write_string(" ", 1);
  }
  void writeString(string s, bool space = true) {
    write_string(s.c_str(), s.size());
    if (space) write_string(" ", 1);
  }
  void writeEndl() { write_string("\n", 1); }
  void flush() {
    fputs(obuf, stdout);
    ocur = 0;
    obuf[0] = 0;
  }

 private:
  bool lflush;

 public:
  FastIO(bool local_flush) {
    obuf[0] = 0;
    lflush = local_flush;
  }
  ~FastIO() { fputs(obuf, stdout); }
};
FastIO<10000000, 10000000> IO(true);
const int maxn = 1e5 + 179;
const int maxi = 11;
const long long MOD = 1e9 + 7;
long long m, k;
vector<int> graph[maxn];
long long dp[maxn][maxi][3];
void dfs(int v, int par) {
  for (int u : graph[v]) {
    if (u != par) dfs(u, v);
  }
  dp[v][0][0] = k - 1;
  dp[v][1][1] = 1;
  dp[v][0][2] = m - k;
  for (int u : graph[v]) {
    if (u != par) {
      long long ndp[maxi][3];
      for (int i = 0; i < ((int)(maxi)); ++i) {
        for (int j = 0; j < ((int)(3)); ++j) ndp[i][j] = 0;
      }
      for (int i = 0; i < ((int)(maxi)); ++i) {
        for (int j = 0; j < ((int)(i + 1)); ++j) {
          ndp[i][0] +=
              dp[v][i - j][0] * (dp[u][j][0] + dp[u][j][1] + dp[u][j][2]);
          ndp[i][0] %= MOD;
          ndp[i][1] += dp[v][i - j][1] * dp[u][j][0];
          ndp[i][1] %= MOD;
          ndp[i][2] += dp[v][i - j][2] * (dp[u][j][0] + dp[u][j][2]);
          ndp[i][2] %= MOD;
        }
      }
      for (int i = 0; i < ((int)(maxi)); ++i) {
        for (int j = 0; j < ((int)(3)); ++j) dp[v][i][j] = ndp[i][j];
      }
    }
  }
}
int main() {
  int n = IO.readInt<int>();
  m = IO.readInt<int>();
  for (int i = 0; i < ((int)(n - 1)); ++i) {
    int u = IO.readInt<int>() - 1;
    int v = IO.readInt<int>() - 1;
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  k = IO.readInt<int>();
  int x = IO.readInt<int>();
  dfs(0, -1);
  long long ans = 0;
  for (int i = 0; i < ((int)(x + 1)); ++i) {
    for (int j = 0; j < ((int)(3)); ++j) ans += dp[0][i][j];
    ans %= MOD;
  }
  IO.writeInt(ans);
  IO.writeEndl();
  return 0;
}