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.*;
import java.math.*;

public class Main{
    
	public static int find(int[] array, int pos) {
		if (array[pos] != pos) {
			return array[pos] = find(array, array[pos]);
		}
		return pos;
	}
 
	public static int pow(int a, int n) {
		long ans = 1;
		long base = a;
		int mod = (int) (1e9 + 7);
		while (n != 0) {
			if ((n & 1) == 1) {
				ans *= base;
				ans %= mod;
			}
			base = (base * base) % mod;
			n >>= 1;
		}
		return (int) ans % mod;
	}
    
	public static void main(String[] args) {
	    
		Scanner s = new Scanner(System.in);
		int mod = (int) (1e9 + 7);
		int vert = s.nextInt();
		int comprimento = s.nextInt();
		int[] array = new int[vert + 1];
		int[] arrayAux = new int[vert + 1];
		for (int i = 1; i <= vert; i++) {
			array[i] = i;
			arrayAux[i] = 1;
		}
		
		for (int i = 1; i < vert; i++) {
			int inicio = s.nextInt();
			int fim = s.nextInt();
			int cor = s.nextInt();
			if (cor == 0) {
				int auxInicio = find(array, inicio);
				int auxFim = find(array, fim);
				array[auxInicio] = auxFim;
				arrayAux[auxFim] += arrayAux[auxInicio];
			}
		}
		
		long resp = 0;
		for (int i = 1; i <= vert; i++) {
			if (array[i] == i) {
				resp = (resp + pow(arrayAux[i], comprimento)) % mod;	
			}
		}
		
		System.out.println(((long) pow(vert, comprimento) - resp + mod) % mod);
	}
}
						 								       				