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 JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.text.*;
public class Main{
    //SOLUTION BEGIN
    //Into the Hardware Mode
    void pre() throws Exception{}
    void solve(int TC)throws Exception {
        int n = ni();
        long m = nl();
        int[] from = new int[n-1], to = new int[n-1];
        for(int i = 0; i< n-1; i++){
            from[i] = ni()-1;
            to[i] = ni()-1;
        }
        int k = ni(), x = ni();
        int[][] g = makeU(n, from, to);
        long[][][] dp = new long[n][1+x][3];//0 -> <k, 1 -> k, 2 -> >k
        dfs(g, dp, 0, -1, k-1, m-k);
        long ans = 0;
        for(int i = 0; i<= x; i++)for(int ty = 0; ty < 3; ty++)ans = (ans+dp[0][i][ty])%MOD;
        pn(ans);
    }
    long MOD = (long)1e9+7;
    void dfs(int[][] g, long[][][] dp, int u, int p, long lo, long hi){
        for(int v:g[u])
            if(v != p)
                dfs(g, dp, v, u, lo, hi);
        int t = 0, x = dp[u].length-1;
        long[][][] ndp = new long[2][1+x][4];
        ndp[t][0][0] = 1;
        for(int v:g[u]){
            if(v == p)continue;
            for(int xx = 0; xx <= x; xx++)for(int ty = 0; ty < 4; ty++)ndp[t^1][xx][ty] = 0;

            for(int x1 = 0; x1 <= x; x1++){
                for(int ty1 = 0; ty1 < 4; ty1++){
                    for(int x2 = 0; x2 <= x-x1; x2++){
                        for(int ty2 = 0; ty2 < 3; ty2++){
                            ndp[t^1][x1+x2][ty1|ty2] += (ndp[t][x1][ty1]*dp[v][x2][ty2])%MOD;
                            if(ndp[t^1][x1+x2][ty1|ty2] >= MOD)ndp[t^1][x1+x2][ty1|ty2] -= MOD;
                        }
                    }
                }
            }
            t ^= 1;
        }
        for(int xx = 0; xx <= x; xx++){
            for(int ty = 0; ty < 4; ty++){
                if(ndp[t][xx][ty] == 0)continue;
                dp[u][xx][0] += (lo*ndp[t][xx][ty])%MOD;

                if(dp[u][xx][0] >= MOD)dp[u][xx][0] -= MOD;
                if(xx < x && (ty&3) == 0){
                    dp[u][xx+1][1] += ndp[t][xx][ty];
                    if(dp[u][xx+1][1] >= MOD)dp[u][xx+1][1] -= MOD;
                }
                if((ty&1)==0){
                    dp[u][xx][2] += (ndp[t][xx][ty]*hi)%MOD;
                    if(dp[u][xx][2] >= MOD)dp[u][xx][2] -= MOD;
                }
            }
        }
    }
    int[][] makeU(int n, int[] from, int[] to){
        int[] cnt = new int[n];
        int[][] g = new int[n][];
        for(int i:from)cnt[i]++;for(int i:to)cnt[i]++;
        for(int i = 0; i< n; i++)g[i] = new int[cnt[i]];
        for(int i = 0; i< n-1; i++){
            g[from[i]][--cnt[from[i]]] = to[i];
            g[to[i]][--cnt[to[i]]] = from[i];
        }
        return g;
    }
    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    void exit(boolean b){if(!b)System.exit(0);}
    long IINF = (long)1e15;
    final int INF = (int)1e9+2, MX = (int)2e6+5;
    DecimalFormat df = new DecimalFormat("0.00000000000");
    double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-7;
    static boolean multipleTC = false, memory = false, fileIO = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        long ct = System.currentTimeMillis();
        if(fileIO){
            in = new FastReader("");
            out = new PrintWriter("");
        }else {
            in = new FastReader();
            out = new PrintWriter(System.out);
        }
        //Solution Credits: Taranpreet Singh
        int T = (multipleTC)?ni():1;
        pre();
        for(int t = 1; t<= T; t++)solve(t);
        out.flush();
        out.close();
        System.err.println(System.currentTimeMillis()-ct);
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }
    int find(int[] set, int u){return set[u] = (set[u] == u?u:find(set, set[u]));}
    int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next() throws Exception{
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }

        String nextLine() throws Exception{
            String str = "";
            try{
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }
            return str;
        }
    }
}