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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class TaskC {

	static int[][] graph;
	static long counter;
	static long mod = 1000000007;
	
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		int n = in.nextInt();
		int k = in.nextInt();
		graph = new int[n][n];
		for (int i = 0; i < n-1; i++) {
			int u = in.nextInt()-1;
			int v = in.nextInt()-1;
			int x = in.nextInt();
			if(x == 0) {
				graph[u][v] = 1;
				graph[v][u] = 1;
			}
		}
		boolean[] visited = new boolean[n];
		long total = 0;
		for (int i = 0; i < n; i++) {
			if(!visited[i]) {
				counter = 0;
				dfs(i, visited);
				counter %= mod;
				total += exponentMod(counter, k);
			}
		}
		total = exponentMod(n, k) - total;
		out.println(total);
		out.close();
	}

	static long exponentMod(long A, long B) {  

		if (A == 0)  
			return 0;  
		if (B == 0)  
			return 1;  

		long y;  
		if (B % 2 == 0){  
			y = exponentMod(A, B / 2);  
			y = (y * y) % mod;  
		}
		else{  
			y = A % mod;  
			y = (y * exponentMod(A, B - 1) % mod) % mod;  
		}  

		return (long)((y + mod) % mod);  
	}   
	
	static void dfs(int v, boolean[] visited) {
		visited[v] = true;
		counter++;
		
		for (int i = 0; i < graph[v].length; i++) {
			if(!visited[i] && graph[v][i] == 1)
				dfs(i, visited);
		}
	}
	
	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

	}

}