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.Scanner;

/* Name of the class has to be "Main" only if the class is public. */

public class q1 {

	static int ctr=1;
	
	public static long pow(int a,int b,int c) {
		
		if(b==0) {
			return 1;
		}
		
		long ans = pow(a,b/2,c);
		
		if(b%2==0) {
			return ((ans%c)*(ans%c))%c;
		}
		else {
			return ((ans%c)*(ans%c)*a)%c;
		}
	}
	public static void connected(boolean a[][],boolean visited[],int start) {
		
		visited[start]=true;
		
		for(int i=0;i<visited.length;i++) {
			if(!visited[i] && a[start][i]) {
				visited[i]=true;
				ctr++;
				connected(a,visited,i);
			}
		}
	}
	
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		int k = s.nextInt();
		boolean a[][] = new boolean[n][n];
		int l=n;
		while(l--!=1) {
			int s1=s.nextInt()-1;
			int s2=s.nextInt()-1;
			int c = s.nextInt();
			if(c==0) {
				a[s1][s2]=a[s2][s1]=true;
			}
		}
		
		boolean visited[] = new boolean[n];
		long sum=0;
		int num = 1000000007;
		for(int i=0;i<n;i++) {
			if(!visited[i]) {
				connected(a,visited,i);
				sum=(sum%num + pow(ctr, k,num))%num;
			}
		}
		
		long total = ((pow(n,k,num))-sum);
		System.out.println(total);
		

	}

	

}