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:
// Main Code at the Bottom
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.BigInteger; 
public class Main {
	//Fast IO class
    static class FastReader {
        BufferedReader br; 
        StringTokenizer st; 
        public FastReader() {
        	boolean env=System.getProperty("ONLINE_JUDGE") != null;
        	if(!env) {
        		try {
					br=new BufferedReader(new FileReader("src\\input.txt"));
				} catch (FileNotFoundException e) {
					e.printStackTrace();
				}
        	}
        	else br = new BufferedReader(new InputStreamReader(System.in)); 
        } 
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) {
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
        int nextInt() {
            return Integer.parseInt(next()); 
        } 
        long nextLong() {
            return Long.parseLong(next()); 
        } 
        double nextDouble() {
            return Double.parseDouble(next()); 
        } 
        String nextLine() {
            String str = ""; 
            try {
                str = br.readLine(); 
            } 
            catch (IOException e) {
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }     
    static long MOD=1000000000+7;
    //Euclidean Algorithm
    static long gcd(long A,long B){
    	return (B==0)?A:gcd(B,A%B);
    }
    //Modular Exponentiation
    static long fastExpo(long x,long n){
        if(n==0) return 1;
        if((n&1)==0) return fastExpo((x*x)%MOD,n/2)%MOD;
        return ((x%MOD)*fastExpo((x*x)%MOD,(n-1)/2))%MOD;
    }
    //Modular Inverse
    static long inverse(long x) {
    	return fastExpo(x,MOD-2);
    }
    //Prime Number Algorithm
    static boolean isPrime(long n){
        if(n<=1) return false;
        if(n<=3) return true;
        if(n%2==0 || n%3==0) return false;
        for(int i=5;i*i<=n;i+=6) if(n%i==0 || n%(i+2)==0) return false;
        return true;
    }
    //Reverse an array
    static void reverse(char arr[],int l,int r){
    	while(l<r) {
    		char tmp=arr[l];
    		arr[l++]=arr[r];
    		arr[r--]=tmp;
    	}
    }
    //Print array
    static void print1d(int arr[]) {
    	out.println(Arrays.toString(arr));
    }
    static void print2d(long arr[][]) {
    	for(long a[]: arr) out.println(Arrays.toString(a));
    }
    // Pair
    static class pair{
    	int x,y;
    	pair(int a,int b){
    		this.x=a;
    		this.y=b;
    	}
    	public boolean equals(Object obj) {
    		if(obj == null || obj.getClass()!= this.getClass()) return false;
            pair p = (pair) obj;
            return (this.x==p.x && this.y==p.y);
        }
    	public int hashCode() {
            return Objects.hash(x,y);
        }
    }
    static FastReader sc=new FastReader();
    static PrintWriter out=new PrintWriter(System.out);  
    //Main function(The main code starts from here)
    static int V,m,k,max;
    static long dp[][][];
    static ArrayList<Integer> graph[];
    static void init(int n){
        V=n;
        graph=new ArrayList[V];
        for(int i=0;i<V;i++) graph[i]=new ArrayList<>();
    }
    static void addEdge(int src,int dest){
        graph[src].add(dest);
        graph[dest].add(src);
    }
    static void dfs(int s,int p) {
    	dp[s][0][0]=k-1;
    	dp[s][1][1]=1;
    	dp[s][0][2]=m-k;
    	for(Integer x: graph[s]) {
    		if(x==p) continue;
    		dfs(x,s);
    		long tmp[][]=new long[max+1][3];
    		for(int i=0;i<=max;i++) {
    			for(int j=0;j+i<=max;j++) {
    				//for type 0
    				tmp[i+j][0]=(tmp[i+j][0]%MOD+(dp[s][i][0]*(dp[x][j][0]+dp[x][j][1]+dp[x][j][2]))%MOD)%MOD;
    				tmp[i+j][1]=(tmp[i+j][1]%MOD+(dp[s][i][1]*dp[x][j][0])%MOD)%MOD;
    				tmp[i+j][2]=(tmp[i+j][2]%MOD+(dp[s][i][2]*(dp[x][j][0]+dp[x][j][2]))%MOD)%MOD;
    			}
    		}
    		for(int i=0;i<=max;i++) for(int t=0;t<3;t++) dp[s][i][t]=tmp[i][t];
    	}
    }
    public static void main (String[] args) throws java.lang.Exception {
    	int test=1;
    	//test=sc.nextInt();
    	while(test-->0){
    		int n=sc.nextInt();m=sc.nextInt();
    		init(n);
    		for(int i=0;i<n-1;i++) addEdge(sc.nextInt()-1,sc.nextInt()-1);
    		k=sc.nextInt();max=sc.nextInt();
    		dp=new long[n][max+1][3];
    		dfs(0,-1);
    		long ans=0;
    		for(int i=0;i<=max;i++) for(int t=0;t<3;t++) ans=(ans%MOD+dp[0][i][t]%MOD)%MOD;
    		out.println(ans%MOD);
    	}
        out.flush();
        out.close();
    }
}