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 solution{
    static Map<Integer,List<Integer>> graph=new HashMap<>();
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int k=sc.nextInt();
        for(int i=0;i<n-1;i++){
            int a=sc.nextInt();
            int b=sc.nextInt();
            int c=sc.nextInt();
            if(c==0) populate(a,b);
        }
        boolean[]visited=new boolean[n+1];
        double total=Math.pow(n,k);
        for(int i=1;i<=n;i++){
            if(!visited[i]){
                double current=dfs(i,visited);
                total-=Math.pow(current,k);
            }
        }
        total=total%(Math.pow(10,9)+7);
        System.out.println(total);
    }
    public static int dfs(int src,boolean[] visited){
        int current=0;
        visited[src]=true;
        for(int nei:graph.getOrDefault(src,new ArrayList<>())){
            if(!visited[nei]){
                current+=dfs(nei,visited);
            }
        }
        return 1+current;
    }
    public static void populate(int i,int j){
        graph.putIfAbsent(i,new ArrayList<>());
        graph.putIfAbsent(j,new ArrayList<>());
        graph.get(i).add(j);
        graph.get(j).add(i);
    }
}