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.ArrayList;
import java.util.Scanner;

public class EdgyTree {
	
	static boolean[] visited;
	
	public static int dfs(ArrayList<ArrayList<Integer>> tree, int idx) {
		int sum = 1;
		visited[idx] = true;
		for(int i=0;i<tree.get(idx).size();i++) {
			int val = tree.get(idx).get(i);
			if(!visited[val])
				sum += dfs(tree,val);
		}
		return sum;
	}
	
	public static void main(String[] args) {
		
		Scanner in = new Scanner(System.in);
		int n,k;
		n = in.nextInt(); k = in.nextInt();
		ArrayList<ArrayList<Integer>> tree = new ArrayList<>();
		for(int i=0;i<n;i++)
			tree.add(new ArrayList<Integer>());
		for(int i=0;i<n-1;i++) {
			int u,v,x;
			u = in.nextInt()-1; v = in.nextInt()-1; x = in.nextInt();
			if(x != 1) {
				tree.get(u).add(v);
				tree.get(v).add(u);
			}
		}
		visited = new boolean[n];
		long res = 0;
		int tmp = 0;
		for(int i=0;i<n;i++) {
			if(!visited[i]) {
				tmp = dfs(tree,i);
				res += Math.pow(tmp, k);
			}
		}
		long tot = (long)Math.pow(n,k);
		System.out.println(((tot - res) % Math.pow(10,9)) + 7);
		in.close();
		
	}
	
}
