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>
#pragma GCC optimize("O3")
using namespace std;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
template <typename T, size_t N>
int SIZE(const T (&t)[N]) {
  return N;
}
template <typename T>
int SIZE(const T &t) {
  return t.size();
}
string to_string(const string s, int x1 = 0, int x2 = 1e9) {
  return '"' + ((x1 < s.size()) ? s.substr(x1, x2 - x1 + 1) : "") + '"';
}
string to_string(const char *s) { return to_string((string)s); }
string to_string(const bool b) { return (b ? "true" : "false"); }
string to_string(const char c) { return string({c}); }
template <size_t N>
string to_string(const bitset<N> &b, int x1 = 0, int x2 = 1e9) {
  string t = "";
  for (int __iii__ = min(x1, SIZE(b)), __jjj__ = min(x2, SIZE(b) - 1);
       __iii__ <= __jjj__; ++__iii__) {
    t += b[__iii__] + '0';
  }
  return '"' + t + '"';
}
template <typename A, typename... C>
string to_string(const A(&v), int x1 = 0, int x2 = 1e9, C... coords);
int l_v_l_v_l = 0, t_a_b_s = 0;
template <typename A, typename B>
string to_string(const pair<A, B> &p) {
  l_v_l_v_l++;
  string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  l_v_l_v_l--;
  return res;
}
template <typename A, typename... C>
string to_string(const A(&v), int x1, int x2, C... coords) {
  int rnk = rank<A>::value;
  string tab(t_a_b_s, ' ');
  string res = "";
  bool first = true;
  if (l_v_l_v_l == 0) res += '\n';
  res += tab + "[";
  x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v));
  auto l = begin(v);
  advance(l, x1);
  auto r = l;
  advance(r, (x2 - x1) + (x2 < SIZE(v)));
  for (auto e = l; e != r; e = next(e)) {
    if (!first) {
      res += ", ";
    }
    first = false;
    l_v_l_v_l++;
    if (e != l) {
      if (rnk > 1) {
        res += '\n';
        t_a_b_s = l_v_l_v_l;
      };
    } else {
      t_a_b_s = 0;
    }
    res += to_string(*e, coords...);
    l_v_l_v_l--;
  }
  res += "]";
  if (l_v_l_v_l == 0) res += '\n';
  return res;
}
void dbgm() { ; }
template <typename Heads, typename... Tails>
void dbgm(Heads H, Tails... T) {
  cout << to_string(H) << " | ";
  dbgm(T...);
}
const long long MOD = (int)1e9 + 7;
const long long INF = 2e18 + 5;
struct hashLL {
  size_t operator()(long long x) const {
    x = (x ^ (x >> 30)) * UINT64_C(0xbf58476d1ce4e5b9);
    x = (x ^ (x >> 27)) * UINT64_C(0x94d049bb133111eb);
    x = x ^ (x >> 31);
    return x;
  }
};
long long inv(long long a, long long b) {
  return 1 < a ? b - inv(b % a, a) * b / a : 1;
}
long long gcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
long long distsq(long long x1, long long y1, long long x2, long long y2) {
  return (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1);
}
const int N = 100001;
long long n, m, u, v, k, x;
unordered_map<long long, vector<int>> g;
long long dp[N][11][3];
long long currdp[11][3];
void dfs(long long node, long long par) {
  dp[node][0][0] = k - 1;
  dp[node][1][1] = 1;
  dp[node][0][2] = m - k;
  for (auto nei : g[node]) {
    if (nei == par) continue;
    dfs(nei, node);
    memset(currdp, 0, sizeof(currdp));
    for (int i = 0; i < (x + 1); i++) {
      for (int j = 0; j < (i + 1); j++) {
        long long ltk = dp[nei][i - j][0];
        long long eqk = dp[nei][i - j][1];
        long long gtk = dp[nei][i - j][2];
        currdp[i][0] += dp[node][j][0] * (ltk + eqk + gtk);
        currdp[i][1] += dp[node][j][1] * (ltk);
        currdp[i][2] += dp[node][j][2] * (ltk + gtk);
        currdp[i][0] %= MOD;
        currdp[i][1] %= MOD;
        currdp[i][2] %= MOD;
      }
    }
    for (int i = 0; i < (x + 1); i++)
      for (int j = 0; j < (3); j++) {
        dp[node][i][j] = currdp[i][j];
      }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  memset(dp, 0, sizeof(dp));
  for (int i = 0; i < (n - 1); i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  cin >> k >> x;
  dfs(1, -1);
  long long ans = 0;
  for (int i = 0; i < (x + 1); i++)
    for (int j = 0; j < (3); j++) {
      ans += dp[1][i][j];
      ans %= MOD;
    }
  cout << ans;
  return 0;
}