You are given a tree (a connected undirected graph without cycles) of n vertices. Each of the n - 1 edges of the tree is colored in either black or red.

You are also given an integer k. Consider sequences of k vertices. Let's call a sequence [a_1, a_2, …, a_k] good if it satisfies the following criterion:

  * We will walk a path (possibly visiting same edge/vertex multiple times) on the tree, starting from a_1 and ending at a_k. 
  * Start at a_1, then go to a_2 using the shortest path between a_1 and a_2, then go to a_3 in a similar way, and so on, until you travel the shortest path between a_{k-1} and a_k.
  * If you walked over at least one black edge during this process, then the sequence is good. 

<image>

Consider the tree on the picture. If k=3 then the following sequences are good: [1, 4, 7], [5, 5, 3] and [2, 3, 7]. The following sequences are not good: [1, 4, 6], [5, 5, 5], [3, 7, 3].

There are n^k sequences of vertices, count how many of them are good. Since this number can be quite large, print it modulo 10^9+7.

Input

The first line contains two integers n and k (2 ≤ n ≤ 10^5, 2 ≤ k ≤ 100), the size of the tree and the length of the vertex sequence.

Each of the next n - 1 lines contains three integers u_i, v_i and x_i (1 ≤ u_i, v_i ≤ n, x_i ∈ \{0, 1\}), where u_i and v_i denote the endpoints of the corresponding edge and x_i is the color of this edge (0 denotes red edge and 1 denotes black edge).

Output

Print the number of good sequences modulo 10^9 + 7.

Examples

Input


4 4
1 2 1
2 3 1
3 4 1


Output


252

Input


4 6
1 2 0
1 3 0
1 4 0


Output


0

Input


3 5
1 2 1
2 3 0


Output


210

Note

In the first example, all sequences (4^4) of length 4 except the following are good: 

  * [1, 1, 1, 1]
  * [2, 2, 2, 2]
  * [3, 3, 3, 3]
  * [4, 4, 4, 4] 



In the second example, all edges are red, hence there aren't any good sequences.
import java.util.*;
public class vtree {

    static  int mod = 1000000007;
    static long pow(int n,int k){
        long ans = 1;

        for(int i=1;i<=k;i++){
            ans = (ans*n)%mod;

        }


        return ans;
    }
    public static  void main(String args[]){

        Scanner s= new Scanner(System.in);
        int n = s.nextInt();
        int k = s.nextInt();
        long ans = 0;
        Dsu dsu = new Dsu(n);
        for(int i=0;i<n-1;i++){
            int x  = s.nextInt();
            int y = s.nextInt();
            int w = s.nextInt();
            if(w == 0){
                dsu.union(x-1,y-1);
            }
        }

        int par[] = dsu.par;
        List<Integer> list = new ArrayList<>();
        for(int i=0;i<n;i++){
            if(par[i]==i){
                list.add(i);
            }
        }
        int size[] = dsu.size;
        ans = pow(n,k) - n;
        for(int x:list){
            int c = 0;
            c = size[x];

            ans = (ans -  (pow(c,k)-c) + mod)%mod;
        }

        System.out.println(ans);
    }

    static class Dsu{

        int n;
        int par[];
        int size[];


        Dsu(int n){
            this.n = n;
            par = new int[n];
            size = new int[n];
            Arrays.fill(size,1);

            for(int i=0;i<n;i++){

                par[i] = i;
            }
        }

        int  find(int x){
            if(par[x]==x){
                return x;
            }
            int tem = find(par[x]);
             par[x] = tem;
            return tem;
        }
        boolean union(int x ,int y){
            int rootx = find(x);
            int rooty = find(y);

            if(rootx == rooty){
                return false;
            }

            if(size[rootx]>=size[rooty]){
                par[rooty] = rootx;
                size[rootx] += size[rooty];

            }
            else{
                par[rootx] = rooty;
                size[rooty] += size[rootx];

            }

          return true;
        }



    }
}
