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.io.*;
import java.util.*;

public class EdgyTrees {
	
	static int MOD = (int)(1e9 + 7);
	public static void main(String[] args) throws Exception {
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		
		StringTokenizer st = new StringTokenizer(f.readLine());
		int n = Integer.parseInt(st.nextToken());
		int k = Integer.parseInt(st.nextToken());
		
		ArrayList<Edge>[] adj = new ArrayList[n];
		
		for(int i = 0; i < n; i ++) {
			adj[i] = new ArrayList<Edge>();
		}
		for(int i = 0; i < n-1; i ++) {
			st = new StringTokenizer(f.readLine());
			int x = Integer.parseInt(st.nextToken()) - 1;
			int y = Integer.parseInt(st.nextToken()) - 1;
			int type = Integer.parseInt(st.nextToken());
			
			adj[x].add(new Edge(y,type));
			adj[y].add(new Edge(x,type));
		}
		
		boolean[] vis = new boolean[n];
		
		long ans = 0;
		for(int i = 0; i < n; i ++) {
			if(vis[i]) continue;
			vis[i] = true;
			Queue<Integer> q = new LinkedList<Integer>();
			q.add(i);
			long count = 1;
			while(!q.isEmpty()) {
				int x = q.remove();
				for(Edge e: adj[x]) {
					if(!vis[e.to] && e.type==0) {
						vis[e.to] = true;
						q.add(e.to);
						count ++;
					}
				}
			}
			ans = (ans + pow(count,k)) % MOD;
		}
		
		System.out.println(((pow(n,k) - ans) % MOD + MOD) % MOD);

	}
	
	static long pow(long n, int k) {
		if(k == 0) return 1;
		return (n * pow(n,k-1)) % MOD;
	}
	
	static class Edge{
		int to;
		int type;
		public Edge(int to, int type) {
			this.to= to;
			this.type = type;
		}
	}

}
