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 sum[] = new int[1000000];
	static int sz[] = new int[1000000];

	public static void main(String[] args) {
		// TODO Auto-generated method stub

		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			int n = sc.nextInt(), k = sc.nextInt();
			for (int i = 1; i <= n; i++) {
				sum[i] = i;sz[i]=0;
			}
			for (int i = 1; i < n; i++) {
				int u = sc.nextInt(), v = sc.nextInt(), x = sc.nextInt();
				if (x == 0) {
					unit(u, v);
				}
			}
			for (int i = 1; i <= n; i++) {
				find(i);
			}
			for (int i = 1; i <= n; i++) {
				sz[sum[i]]++;
			}
			long p = 0;
			for (int i = 1; i <= n; i++) {
				p += poww(sz[i], k);
				p%=1000000007;
			}
			long an = (poww(n, k) + 1000000007 - p) % 1000000007;
			System.out.println(an);
		}
	}

	public static long poww(int x, int n) {// 快速幂
		if (n == 0)
			return 1;
		long t = poww(x, n / 2) % 1000000007;
		t = t * t;
		t %= 1000000007;
		if (n % 2 == 1) {
			t = t * x;
			t %= 1000000007;
		}
		return t;
	}

	static int find(int x) {
		if (sum[x] == x) {
			return x;
		} else {
			sum[x] = find(sum[x]);
			return sum[x];
		}
	}

//合并两子集的函数
	static void unit(int x, int y) {// 获取xy合并
		int fx = find(x);// 找到各自的老大
		int fy = find(y);
		if (fx != fy) {// 如果各自老大不一样
			sum[fy] = fx;// 就靠左
		}
		return;// 加的
	}
}
