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  Edgy_Trees
{
	public static void main(String[] args) throws IOException 
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw=new PrintWriter(System.out);

		int t=1;

		while(t-->0)
		{ 
			String str[]=br.readLine().trim().split(" ");
			int n=Integer.parseInt(str[0]);
			int k=Integer.parseInt(str[1]);

			adj=new ArrayList[n+1];
			for(int i=0;i<n+1;i++)
				adj[i]=new ArrayList<>();

			for(int i=0;i<n-1;i++)
			{
				str=br.readLine().trim().split(" ");

				int type=Integer.parseInt(str[2]);

				int x=Integer.parseInt(str[0]);
				int y=Integer.parseInt(str[1]);

				if(type==0)
				{
					adj[x].add(y);
					adj[y].add(x);
				}
			}

			v=new boolean[n+1];
			int ans=0;
			int toal_node=0;
			long mod=1000000007;
			for(int i=1;i<n+1;i++)
			{
				if(!v[i])
				{
					count=0;
					dsf(i,-1);
					toal_node+=count;
					ans+=power(count,(long)k,mod);
					//pw.println(count);
				}
			}


			long total=power((long)n,(long)k,mod);

			pw.println((total-ans+mod)%mod);


			
		}
		pw.flush();
	}

	static ArrayList<Integer> adj[];
	static int n;
	static int k;

	static int count;
	static boolean v[];

	public static void dsf(int c,int p)
	{
		if(v[c])
			return;

		v[c]=true;
		count++;
		for(int x:adj[c])
		{
			if(x!=p && !v[x])
				dsf(x,c);
		}
	}

	static long power(long x, long y, long m)  
    { 
        if (y == 0) 
            return 1; 
          
        long p = power(x, y / 2, m) % m; 
        p = (p * p) % m; 
      
        if (y % 2 == 0) 
            return p; 
        else
            return (x * p) % m; 
    } 
}