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.
// package cp;
import java.io.*;
import java.util.*;
public class Cf_three {
	public static void main(String[] args) throws IOException {	
		PrintWriter out = new PrintWriter(System.out);
		Readers.init(System.in);
		int n=Readers.nextInt();
		int k=Readers.nextInt();
		int[] a=new int[n+1];
		int cnt=1;
		for (int i = 0; i < n-1; i++) {
			int u=Readers.nextInt();
			int v=Readers.nextInt();
			int x=Readers.nextInt();
			if(x==0) {
				if(a[u]==0 && a[v]==0) {
					a[u]=cnt;
					a[v]=cnt;
					cnt++;
				}
				else {
					a[u]=Math.max(a[u], a[v]);
					a[v]=Math.max(a[u], a[v]);
				}
			}
		}
		HashMap<Integer,Long> map=new HashMap<>();
		for (int i = 1; i <=n; i++) {
			if(map.get(a[i])==null) {
				map.put(a[i],(long) 1);
			}
			else map.put(a[i],map.get(a[i])+1);
		}
		long zero=0;
		long summ=0;
		for(Integer temp:map.keySet()) {
			if(temp==0)zero=map.get(temp);
			else summ=(long) (summ+Math.pow(map.get(temp), k));
		}
		long ans=(long) (Math.pow(n, k)-summ-zero)%(1000000007);
		System.out.println(ans);
		out.flush();
	}

}
class Readers {

    static BufferedReader reader;

    static StringTokenizer tokenizer;



    /** call this method to initialize reader for InputStream */

    static void init(InputStream input) {

        reader = new BufferedReader(

                     new InputStreamReader(input) );

        tokenizer = new StringTokenizer("");

    }



    /** get next word */

    static String next() throws IOException {

        while ( ! tokenizer.hasMoreTokens() ) {

            //TODO add check for eof if necessary

            tokenizer = new StringTokenizer(

                   reader.readLine() );

        }

        return tokenizer.nextToken();

    }



    static int nextInt() throws IOException {

        return Integer.parseInt( next() );

    }

	

    static double nextDouble() throws IOException {

        return Double.parseDouble( next() );

    }
    
    
    static long nextLong() throws IOException{
    	
		return Long.parseLong(next());
	}

}