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.ArrayList;
import java.util.HashMap;
import java.util.Map.Entry;
import java.util.Scanner;



public class P1139C {

	static Scanner in;
	
	static int n, k;
	static HashMap<Integer, ArrayList<Edge>> g;
	static boolean[] marked;
	static int[] id;
	
	public static void main(String[] args) {
		in = new Scanner(System.in);
		
		
		n = in.nextInt();
		k = in.nextInt();

		marked = new boolean[n];
		id = new int[n];
		
		g = new HashMap<>();
		
		for(int i=0; i<n; i++) g.put(i, new ArrayList<Edge>());
		
		int cc=0;
		
		for(int i=0; i<n-1; i++) {
			int u = in.nextInt()-1;
			int v = in.nextInt()-1;
			boolean c = (in.nextInt()==1?true:false);
			
			g.get(u).add(new Edge(v, c));
			g.get(v).add(new Edge(u, c));
		}
		
		for(int i=0; i<n; i++) {
			if (!marked[i]) {
				dfs(i, cc);
				cc++;
			}
		}
		
		long sum=0;
		
		HashMap<Integer, Integer> noOfCC = new HashMap<>();
		
		for(int i=0; i<n; i++) {
			try {
				noOfCC.put(id[i], noOfCC.get(id[i])+1);
			} catch(Exception e) {
				noOfCC.put(id[i], 1);
			}
		}
		
		for(Entry e: noOfCC.entrySet()) {
			
			sum+=(long)Math.pow((Integer)e.getValue(), k);
		}
		
		sum=(long) (Math.pow(n, k)-sum);
		
		System.out.println(sum);

	}
	
	public static void dfs(int v, int cc) {
		marked[v]=true;
		id[v]=cc;
		
		
		for(Edge e: g.get(v)) {
			if (!marked[e.v] && !e.c) {
				dfs(e.v, cc);
			}
		}
	}

	static class Edge{
		int v;
		boolean c;
		
		public Edge(int v, boolean c) {
			this.v=v;
			this.c=c;
		}
		
	}
}
