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.Arrays;
import java.util.HashSet;
import java.util.Scanner;

public class Main {

	private static int n, k;
	private static int[][] edges;
	private static int[] id;
	private static int[] size;
	private static int components;
	private static final int MOD = (int)1e9+7;
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		n = in.nextInt();
		k = in.nextInt();
		edges = new int[n-1][3];
		for(int i=0; i<n-1; i++) {
			int p = in.nextInt()-1;
			int q = in.nextInt()-1;
			int w = in.nextInt();
			edges[i] = new int[] {p, q, w};
		}
		
		id = new int[n];
		size = new int[n];
		
		Arrays.fill(size, 1);
		for(int i=0; i<n; i++) {
			id[i] = i;
		}
		
		for(int i=0; i<n-1; i++) {
			if (edges[i][2]==1) continue;
			// else red edges
			
			union(edges[i][0], edges[i][1]);
		}
		
		
		boolean[] isRoot = new boolean[n];
		
		for(int i=0; i<n; i++) {
			isRoot[root(i)] = true;
		}
		
		long ans = exp(n, k);
		
		for(int i=0; i<n; i++) {
			if (isRoot[i]) {
				ans = (ans-exp(size[i], k)+MOD)%MOD;
			}
		}
		System.out.println(ans);
		
		
		
		

	}
	
	
	private static long exp(int n, int k) {
		if (k==0) return 1;
		
		if (k%2!=0) {
			return (n*exp(n, k-1))%MOD;
		} else {
			long halfExp = exp(n, k/2);
			return (halfExp*halfExp)%MOD;
		}
	}

	private static void union(int i, int j) {
		int p = root(i);
		int q = root(j);
		
		if (size[p]>size[q]) {
			id[q] = p;
			size[p] += size[q];
		} else {
			id[p] = q;
			size[q] += size[p];
		}
	}
	
	
	private static int root(int u) {
		while(u!=id[u]) {
			id[u] = id[id[u]]; //path compression
			u = id[u];
		}
		return u;
	}
	

}
