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.LinkedList;
import java.util.Scanner;
 
public class Main {
	
	public static ArrayList<Integer>[] g;
	public static boolean [] visited;
	
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		
		g = new ArrayList[n+1];
		
		for(int i = 1; i <= n; i++) {
			g[i] = new ArrayList<Integer>();
		}
		visited = new boolean[n+1];

		for(int i = 1; i < n; i++) {
			int u = sc.nextInt();
			int v = sc.nextInt();
			int c = sc.nextInt();
			
			if(c != 1) {
				g[u].add(v);
				g[v].add(u);
			}


		}
		
		int res = 0;
		for(int i = 1; i <= n; i++) {
			int p = 0;
			if(!visited[i]) {
				LinkedList<Integer> s = new LinkedList<Integer>();
				s.addFirst(i);
				visited[i] = true;
				p++;
				
				while (!s.isEmpty()) {
					
					int curr = s.removeFirst();
					
					if(!visited[curr]) {
						visited[curr] = true;
						p++;
						
					}
					
					for(int v : g[curr]) {
						if(!visited[v]) {
							s.addFirst(v);
							visited[v] = true;
							p++;
						}
						
					}
				}

			}
			
			if(p != 0) {
				res+=Math.pow(p, k);
			}
			
			
			
		}
		
		res = (int)(Math.pow(n, k)-res)%1000000007;
		System.out.println(res);
				
	}


}