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;

public class cf2 {

	public static void main(String[] args) {
		Scanner scn = new Scanner(System.in);
		int n = scn.nextInt();
		int k = scn.nextInt();
		int[] set = new int[n];
		int[] arr = new int[n];
		for (int i = 0; i < n; i++) {
			set[i] = i;
			arr[i] = 1;
		}

		for (int i = 0; i < n - 1; i++) {
			int x = scn.nextInt() - 1;
			int y = scn.nextInt() - 1;
			int s = scn.nextInt();
			if (s == 0) {
				int fa = find(x, set);
				int fb = find(y, set);
				if (fa != fb) {
					if (arr[fa] > arr[fb]) {
						arr[fa] += arr[fb];
						set[fb] = fa;
					} else {
						arr[fb] += arr[fa];
						set[fa] = fb;
					}

				}
			}

		}
		int ans = 0;
		int mod = 1000000007;
		for (int i = 0; i < n; i++) {
			if (set[i] == i) {
				ans = (int) ((ans + Math.pow(arr[i], k)) % mod);
			}
		}
		System.out.println((int)(Math.pow(n, k) - ans + mod) % mod);
	}

	public static int find(int myindex, int[] set) {

		if (myindex != set[myindex]) {
			return find(set[myindex], set);
		} else {
			return myindex;
		}
	}
}
