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.*;
import java.io.*;
public class Solution{
    static ArrayList<Integer> a[];
    static boolean[] v;
    static class pair implements Comparable<pair>{
        int a,b;
        pair(int x,int y){
            a=x;b=y;
        }
        public int compareTo(pair t){
            if(t.a==this.a)
            return this.b-t.b;
            return this.a-t.a;
        }
    }
    public static ArrayList<pair> bfs(String[] a,int r,int c){
        ArrayList<pair> ans=new ArrayList<>();
        Queue<pair> q=new LinkedList<>();
        int[][] dxy={{-1,0,1,0},{0,-1,0,1}};
        q.add(new pair(r,c));
        HashSet<String> h=new HashSet<>();
        while(!q.isEmpty()){
            pair f=q.poll();
            ans.add(f);h.add(f.a+" "+f.b);
            for(int i=0;i<4;i++){
                int dx=f.a+dxy[0][i];
                int dy=f.b+dxy[1][i];
                if(dx<0||dy<0||dx>=a.length||dy>=a.length||h.contains(dx+" "+dy)||
                a[dx].charAt(dy)=='1')
                continue;
                q.add(new pair(dx,dy));
                h.add(dx+" "+dy);
            }
        }
        return ans;
    }    
    public static long dfs(int a[][],boolean[] v,int n){
        v[n]=true;
        long q=1;
        for(int i=0;i<a.length;i++){
            if(!v[i]&&a[n][i]!=0)
            q+=dfs(a,v,i);
        }
        return q;
    }
    public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
// 		int t = s.nextInt();
// 		while(t-- > 0) {
 
			int n = s.nextInt();
			int k = s.nextInt();
            int[][]g=new int[n][n];
            
            for(int i=0;i<n-1;i++){
                int u=s.nextInt()-1;
                int v=s.nextInt()-1;
                int w=s.nextInt()-1;
                g[u][v]=w;
                g[v][u]=w;
            }
            boolean[] v=new boolean[n];
            long ans=(long)Math.pow(n,k)%1000000007;
            for(int i=0;i<n;i++){
                if(!v[i])
                ans=(ans-(long)Math.pow(dfs(g,v,i),k))%1000000007;
            }
            System.out.println(ans%1000000007);
      
	//	}
	}
}