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.io.*;
import java.util.*;

public class C{
	// ------------------------
	static HashMap<Integer,HashSet<Integer>>tree=new HashMap<Integer,HashSet<Integer>>();
	static boolean[] used;
	static final long MOD=1000000007;
	public static void main(String[] args) {
		MyScanner sc = new MyScanner();
		out = new PrintWriter(new BufferedOutputStream(System.out));
		// ------------------------
		int n=sc.nextInt(),k=sc.nextInt();
		for(int i=0;i<n;i++)
			tree.put(i,new HashSet<Integer>());
		used=new boolean[n];
		for(int i=0;i<n-1;i++){
			int u=sc.nextInt()-1,v=sc.nextInt()-1,x=sc.nextInt();
			if(x==0){
				tree.get(u).add(v);
				tree.get(v).add(u);
			}
		}
		long bad=0;
		for(int i=0;i<n;i++)
			if(!used[i])
				bad+=pow(dfs(i),k);
		long ans=pow(n,k)-bad;
		while(ans<0)
			ans+=MOD;
		System.out.println(ans%MOD);
		// ------------------------
		out.close();
	}
	//------------------------
	private static long pow(long x,long p){
		if(p==0)
			return 1;
		long a=pow(x,p/2);
		return (((a*a)%MOD)*(p%2==1?x:1))%MOD;
	}
	private static long dfs(int i){
		used[i]=true;
		int ans=1;
		for(int j:tree.get(i))
			if(!used[j])
				ans+=dfs(j);
		return ans;
	}
	public static PrintWriter out;
	public static class MyScanner {
		BufferedReader br;
		StringTokenizer st;
		public MyScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
}
