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 Main {
	static int mod = (int) (1e9 + 7);
	static int[] set;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			int n = sc.nextInt();
			int k = sc.nextInt();
			set = new int[n + 1];
			int[] arr = new int[n + 1];
			for (int i = 1; i <= n; i++) {
				set[i] = i;
				arr[i] = 1;
			}
			for (int i = 1; i < n; i++) {
				int a = sc.nextInt();
				int b = sc.nextInt();
				int ha = sc.nextInt();
				if (ha == 0) {
					int fa = find(a);
					int fb = find(b);
					set[fa] = fb;
					arr[fb] += arr[fa];
				}
			}
			long ans = 0;
			for (int i = 1; i <= n; i++) {
				if (set[i] == i) {
					ans = ((long) ans + pow(arr[i], k)) % mod;
				}
			}
			System.out.println(((long) pow(n, k) - ans)%mod);
		}
	}

	public static int find(int x) {
		if (set[x] != x) {
			return set[x] = find(set[x]);
		}
		return x;
	}

	public static int pow(int a, int n) {
		long ans = 1;
		long base = a;
		while (n != 0) {
			if ((n & 1) == 1) {
				ans *= base;
				ans %= mod;
			}
			base = (base * base) % mod;
			n >>= 1;
		}
		return (int) ans % mod;
	}
}
