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>
struct Range {
  struct It {
    int num, step;
    int operator*() { return num; }
    void operator++() { num += step; }
    bool operator!=(const It& other) { return num != other.num; }
  };
  Range(int ee) : b(0), e(ee) {}
  Range(int bb, int ee) : b(bb), e(ee) {}
  It begin() { return {b, (b < e ? 1 : -1)}; }
  It end() { return {e, 0}; }
  int b, e;
};
template <typename T>
inline T& UMAX(T& x, T y) {
  if (x < y) x = y;
  return x;
}
template <typename T>
inline T& UMIN(T& x, T y) {
  if (y < x) x = y;
  return x;
}
template <typename T, typename... Args>
struct ArithmiticPromotion {
  typedef decltype(T() + typename ArithmiticPromotion<Args...>::type()) type;
};
template <typename T, typename U>
struct ArithmiticPromotion<T, U> {
  typedef decltype(T() + U()) type;
};
template <typename T>
struct ArithmiticPromotion<T, T> {};
template <typename T>
struct ArithmiticPromotion<T> {};
template <typename T, typename U>
typename ArithmiticPromotion<T, U>::T MAX(T a, U b) {
  return a < b ? b : a;
}
template <typename T, typename... Args>
typename ArithmiticPromotion<T, Args...>::T MAX(T a, Args... args) {
  return MAX(a, MAX(args...));
}
template <typename T, typename U>
typename ArithmiticPromotion<T, U>::T MIN(T a, U b) {
  return a < b ? a : b;
}
template <typename T, typename... Args>
typename ArithmiticPromotion<T, Args...>::T MIN(T a, Args... args) {
  return MIN(a, MIN(args...));
}
template <typename T>
struct ScanfSpecifier {};
template <>
struct ScanfSpecifier<char*> {
  static constexpr const char* value = "%s";
};
template <>
struct ScanfSpecifier<int> {
  static constexpr const char* value = "%d";
};
template <>
struct ScanfSpecifier<double> {
  static constexpr const char* value = "%lf";
};
template <>
struct ScanfSpecifier<float> {
  static constexpr const char* value = "%f";
};
template <>
struct ScanfSpecifier<char> {
  static constexpr const char* value = "%c";
};
template <>
struct ScanfSpecifier<const char*> {
  static constexpr const char* value = "%s";
};
template <>
struct ScanfSpecifier<unsigned long> {
  static constexpr const char* value = "%lu";
};
template <>
struct ScanfSpecifier<unsigned int> {
  static constexpr const char* value = "%u";
};
template <>
struct ScanfSpecifier<long long int> {
  static constexpr const char* value = "%lld";
};
template <typename T>
int RD(T& arg) {
  return std::scanf(ScanfSpecifier<T>::value, &arg);
}
template <int S>
int RD(char (&arg)[S]) {
  return std::scanf("%s", arg);
}
int RD(char* arg) { return std::scanf("%s", arg); }
template <>
int RD<char>(char& arg) {
  return std::scanf(" %c", &arg);
}
template <typename T, typename... Args>
int RD(T& arg1, Args&... args) {
  return RD(arg1) + RD(args...);
}
template <typename T>
T RD() {
  T ret;
  RD(ret);
  return ret;
}
template <typename It>
void RDV(It begin, It end) {
  while (begin != end) RD(*begin++);
}
template <typename C>
void RDV(C& c) {
  RDV(std::begin(c), std::end(c));
}
template <typename T>
void WT(T arg) {
  std::printf(ScanfSpecifier<T>::value, arg);
}
template <typename T, typename U>
void WT(std::pair<T, U> arg) {
  std::printf("(");
  WT(arg.first);
  std::printf(", ");
  WT(arg.second);
  std::printf(")");
}
template <typename... Args>
void WT(Args... args) {
  int alc = 0;
  int dummy[] = {((alc++ ? std::printf(" ") : 0), WT(args), 0)...};
}
template <typename... Args>
void WTL(Args... args) {
  WT(args...);
  std::printf("\n");
}
template <typename It>
void WTV(It begin, It end) {
  int alc = 0;
  while (begin != end) (alc++ ? std::printf(" ") : 0), WT(*begin++);
}
template <typename C>
void WTV(const C& c) {
  WTV(std::begin(c), std::end(c));
}
template <typename It>
void WTVL(It begin, It end) {
  WTV(begin, end);
  std::printf("\n");
}
template <typename C>
void WTVL(const C& c) {
  WTVL(std::begin(c), std::end(c));
}
template <unsigned int Z>
struct ZpZ {
  static const unsigned int P = Z;
  using U64 = unsigned long long int;
  U64 value;
  ZpZ(long long int n) {
    if (std::abs(n) >= P) n %= P;
    if (n < 0) n += P;
    value = n;
  }
  ZpZ() : value(0) {}
  ZpZ& operator+=(ZpZ p) {
    if ((value += p.value) >= P) value -= P;
    return *this;
  }
  ZpZ& operator-=(ZpZ p) {
    if (value >= p.value)
      value -= p.value;
    else
      value += P - p.value;
    return *this;
  }
  ZpZ& operator*=(ZpZ p) {
    value = value * p.value % P;
    return *this;
  }
  ZpZ& operator/=(ZpZ p) {
    value = value * p.inverse().value % P;
    return *this;
  }
  bool operator==(ZpZ p) { return value == p.value; }
  bool operator!=(ZpZ p) { return value != p.value; }
  ZpZ operator+(ZpZ r) const { return ZpZ(value) += r; }
  ZpZ operator-(ZpZ r) const { return ZpZ(value) -= r; }
  ZpZ operator*(ZpZ r) const { return ZpZ(value) *= r; }
  ZpZ operator/(ZpZ r) const { return ZpZ(value) /= r; }
  explicit operator long long int() const { return value; }
  explicit operator int() const { return value; }
  ZpZ operator-() { return P - value; }
  ZpZ inverse() {
    int a = value, b = P, u = 1, v = 0;
    while (b) {
      int t = a / b;
      std::swap(a -= t * b, b);
      std::swap(u -= t * v, v);
    }
    if (u < 0) u += P;
    return u;
  }
  ZpZ pow(U64 e) {
    U64 ret = 1, base = value;
    while (e) {
      if ((e & 1) && ((ret *= base) >> 32)) ret %= P;
      if ((base *= base) >> 32) base %= P;
      e >>= 1;
    }
    return ret;
  }
};
template <unsigned int P>
void WT(ZpZ<P> arg) {
  WT((int)arg);
}
const int MOD = 1e9 + 7;
using ZZ = ZpZ<MOD>;
ZZ min(ZZ a, ZZ b) { return std::min(a.value, b.value); }
ZZ max(ZZ a, ZZ b) { return std::max(a.value, b.value); }
using RG = Range;
using namespace std;
ZZ dp[100009][11][3];
vector<int> adj[100009];
int N, M;
int K, X;
void f(int u, int p) {
  for (int to : adj[u])
    if (to != p) f(to, u);
  ZZ now[11] = {};
  now[0] = K - 1;
  for (int to : adj[u])
    if (to != p) {
      ZZ tmp[11] = {};
      for (int i : RG(X + 1))
        for (int j : RG(X + 1))
          if (i + j <= X)
            tmp[i + j] += now[i] * (dp[to][j][0] + dp[to][j][1] + dp[to][j][2]);
      for (int i : RG(X + 1)) now[i] = tmp[i];
    }
  for (int i : RG(X + 1)) dp[u][i][0] = now[i];
  memset(now, 0, sizeof(now));
  now[1] = 1;
  for (int to : adj[u])
    if (to != p) {
      ZZ tmp[11] = {};
      for (int i : RG(X + 1))
        for (int j : RG(X + 1))
          if (i + j <= X) tmp[i + j] += now[i] * dp[to][j][0];
      for (int i : RG(X + 1)) now[i] = tmp[i];
    }
  for (int i : RG(X + 1)) dp[u][i][1] = now[i];
  memset(now, 0, sizeof(now));
  now[0] = M - K;
  for (int to : adj[u])
    if (to != p) {
      ZZ tmp[11] = {};
      for (int i : RG(X + 1))
        for (int j : RG(X + 1))
          if (i + j <= X) tmp[i + j] += now[i] * (dp[to][j][0] + dp[to][j][2]);
      for (int i : RG(X + 1)) now[i] = tmp[i];
    }
  for (int i : RG(X + 1)) dp[u][i][2] = now[i];
}
int main() {
  RD(N, M);
  for (int i : RG(N - 1)) {
    int u, v;
    RD(u, v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  RD(K, X);
  f(1, 0);
  ZZ ans = 0;
  for (int i : RG(X + 1)) ans += dp[1][i][0] + dp[1][i][1] + dp[1][i][2];
  WTL(ans);
}