Manthan, Codefest 17 C. Helga Hufflepuff's Cup（樹形DP）

 

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

 

 

 

題意：給你一棵樹，可以染m種顏色，現在定義一種最高值k，一棵樹上最多能有x個最高值，如果一個節點爲最高值k，那麼他相鄰的節點的值只能選比他小的。現在問你一共有多少種染色的方法。

 

思路：dp [i] [j] [k] 表示以i節點爲根的子樹上選j個最高值有多少種方法，k = 0表示選比K小的值，k = 1表示選值爲k，k = 2表示選比k大的值

 

 

#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long

using namespace std;

const int mod = 1e9 + 7;


int n,m,K,X;
vector<int> e[100010];
LL dp[100010][12][3];
int sz[100010];
int t[12][3];

LL dfs(int x,int pre)
{
    dp[x][0][0] = K-1;
    dp[x][1][1] = 1;
    dp[x][0][2] = m-K;
    sz[x] = 1;
    for(int i=0;i<e[x].size();i++)
    {
        int xx = e[x][i];
        if(xx == pre)
            continue;
        dfs(xx,x);
        memset(t,0,sizeof(t));
        for(int j=0;j<=sz[x];j++)
            for(int k=0;k<=sz[xx];k++)
            {
                if(j+k > X)
                    continue;
                t[j+k][0] = (t[j+k][0] + dp[x][j][0]*(dp[xx][k][0] + dp[xx][k][1] + dp[xx][k][2])%mod)%mod;
                t[j+k][1] = (t[j+k][1] + dp[x][j][1]*(dp[xx][k][0])%mod)%mod;
                t[j+k][2] = (t[j+k][2] + dp[x][j][2]*(dp[xx][k][0] + dp[xx][k][2])%mod)%mod;
            }
        sz[x] = min(sz[x]+sz[xx],X);
        for(int j=0;j<=sz[x];j++)
            for(int k=0;k<3;k++)
                dp[x][j][k] = t[j][k];
    }
}
int main(void)
{
    int i,j,k;
    while(scanf("%d%d",&n,&m)==2)
    {
        for(i=1;i<=n;i++)
            e[i].clear();
        for(i=1;i<n;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            e[x].push_back(y);
            e[y].push_back(x);
        }
        scanf("%d%d",&K,&X);
        dfs(1,-1);
        LL ans = 0;
        for(j=0;j<=X;j++)
            for(k=0;k<3;k++)
                ans = (ans + dp[1][j][k])%mod;
        cout << ans << endl;
    }
}