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.LinkedList;
import java.util.Scanner;

public class Graph {
	int V;
	LinkedList<Integer>[] adjListArray;

	Graph(int V) {
		this.V = V;
		adjListArray = new LinkedList[V];

		for (int i = 0; i < V; i++) {
			adjListArray[i] = new LinkedList<Integer>();
		}
	}

	void addEdge(int src, int dest) {
		adjListArray[src].add(dest);
		adjListArray[dest].add(src);
	}

	void DFSUtil(int v, boolean[] visited) {
		visited[v] = true;
		count++;
		for (int x : adjListArray[v]) {
			if (!visited[x])
				DFSUtil(x, visited);
		}

	}
	
	 static int power(int x, int y, int p) 
	    { 
	        // Initialize result 
	        int res = 1;      
	         
	        // Update x if it is more   
	        // than or equal to p 
	        x = x % p;  
	      
	        while (y > 0) 
	        { 
	            // If y is odd, multiply x 
	            // with result 
	            if((y & 1)==1) 
	                res = (res * x) % p; 
	      
	            // y must be even now 
	            // y = y / 2 
	            y = y >> 1;  
	            x = (x * x) % p;  
	        } 
	        return res; 
	    } 

	void connectedComponents() {
		boolean[] visited = new boolean[V];
		for (int v = 0; v < V; ++v) {
			if (!visited[v]) {
				count = 0;
				DFSUtil(v, visited);
				nhihonge = (nhihonge%MOD + power(count,k,MOD)%MOD)%MOD;
			}
		}
	}

	public static int count = 0;
	public static long nhihonge = 0;
	public static int k;
	public static final int MOD = 1000000007;

	public static void main(String[] args) {
		Scanner scn = new Scanner(System.in);
		int n = scn.nextInt();
		k = scn.nextInt();
		Graph g = new Graph(n);
		for (int i = 0; i < n - 1; ++i) {
			int u = scn.nextInt();
			int v = scn.nextInt();
			int x = scn.nextInt();
			if (x == 0) {
				g.addEdge(u - 1, v - 1);
			}
		}
		g.connectedComponents();

		System.out.println((int) ((power(n,k,MOD)%MOD - nhihonge%MOD) % MOD));
	}
}