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.*;
import java.math.BigInteger;

public class Main
{
    static final long mod=(int)1e9+7;
    static ArrayList<pair>[] adj;
    static boolean[] visited;
    public static void main(String[] args) throws Exception
    {
    	FastReader in=new FastReader();
    	PrintWriter pw=new PrintWriter(System.out);
    	int n=in.nextInt();
    	int k=in.nextInt();
    	adj=new ArrayList[n+1];
    	visited=new boolean[n+1];
    	for(int i=1;i<=n;i++)
    		adj[i]=new ArrayList();
    	for(int i=1;i<n;i++)
    	{
    		int a=in.nextInt();
    		int b=in.nextInt();
    		int c=in.nextInt();
    		adj[a].add(new pair(b,c));
    		adj[b].add(new pair(a,c));
    	}
    	long ans=fastMod((long)n,k);
    	for(int i=1;i<=n;i++)
    	{
    		long a=dfs(i);
    		ans-=fastMod(a,k);
    		while(ans<0)
    			ans+=mod;
    	}
    	pw.println(ans);
    	pw.flush();
    }

    static long dfs(int a)
    {
    	if(visited[a])
    		return 0;
    	visited[a]=true;
    	long r=1;
    	for(pair p:adj[a])
    	{
    		if(p.e==0)
    			r+=dfs(p.v);
    	}

    	return r;
    }

    static long fastMod(long n,int k)
    {
    	long r=1;

    	while(k>0)
    	{
    		if((k&1)==1)
    		{
    			r=(r*n)%mod;
    		}
    		n=(n*n)%mod;
    		k>>=1;
    	}

    	return r;
    }
}

class pair
{
	int v,e;

	public pair(int a,int b)
	{
		v=a;
		e=b;
	}
}

class FastReader
{
    BufferedReader br;
    StringTokenizer st;
 
    public FastReader()
    {
        br=new BufferedReader(new InputStreamReader(System.in));
    }
 
    public String next() throws IOException
    {
        if(st==null || !st.hasMoreElements())
        {
            st=new StringTokenizer(br.readLine());
        }
        return st.nextToken();
    }
 
    public int nextInt() throws IOException
    {
        return Integer.parseInt(next());
    }
 
    public long nextLong() throws IOException
    {
        return Long.parseLong(next());
    }
}