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 Problem {

	static int find(int[] parents, int i) {
		if (parents[i] != i) {
			parents[i] = find(parents, parents[i]);
		}
		return parents[i];
	}

	static void union(int[] parents, int[] size, int i, int j) {
		int setI = find(parents, i);
		int setJ = find(parents, j);
		if (setI == setJ)
			return;

		if (size[setI] >= size[setJ]) {
			parents[setJ] = setI;
			size[setI] += size[setJ];
		} else {
			parents[setI] = setJ;
			size[setJ] += size[setI];
		}
	}

	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		String[] next = in.readLine().trim().split(" ");
		int numNodes = Integer.parseInt(next[0]);
		int length = Integer.parseInt(next[1]);
		int[] parents = new int[numNodes];
		int[] size = new int[numNodes];
		long res = (long) Math.pow((double) numNodes, (double) length);
		if(numNodes == 19 && length==20) {
			System.out.println(res);
		}

		for (int i = 0; i < numNodes; i++) {
			parents[i] = i;
			size[i] = 1;
		}
		for (int i = 0; i < numNodes - 1; i++) {
			String[] edge = in.readLine().trim().split(" ");
			if (edge[2].equals("0")) {
				int src =Integer.parseInt(edge[0])-1;
				int dest = Integer.parseInt(edge[1])-1;
				union(parents, size,src,dest);
	
				
			}

		}
		for(int i=0;i<numNodes;i++) {
			
			if(find(parents,i)==i) {			
				res -= (long)Math.pow((double)size[i], (double)length);
			}
		}
		
		System.out.println(res % 1000000007);

		in.close();

	}

}