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.*;
import java.io.*;

public class C {
	static BufferedReader br;
	static StringTokenizer st;
	
	static int n;
	static int k;
	
	static List<Edge> edges;
	
	public static void main(String[] args) throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		st = new StringTokenizer(br.readLine());
		n = Integer.parseInt(st.nextToken());
		k = Integer.parseInt(st.nextToken());
		edges = new ArrayList<>();
		for (int i = 0; i<n-1; i++) {
			st = new StringTokenizer(br.readLine());
			edges.add(new Edge(Integer.parseInt(st.nextToken()), Integer.parseInt(st.nextToken()), Integer.parseInt(st.nextToken())));
		}
		solve();
	}
	
	public static void solve() {
		List<Integer>[] blackless = new ArrayList[n+1];
		for (int i =0 ; i<=n; i++) {
			blackless[i] = new ArrayList<>();
		}
		for (Edge e : edges) {
			if (e.color==0) {
				blackless[e.node1].add(e.node2);
				blackless[e.node2].add(e.node1);
			}
		}
		boolean[] visited = new boolean[n+1];
		long ptok = 0;
		long temp = 1;
		for (int i = 1; i<=n; i++) {
			if (!visited[i]) {
				long tbm = bfs(blackless, i, visited);
				for (int j = 0; j<k; j++) {
					temp*=tbm;
					temp%=(long)(10e9+7);
				}
				ptok+=temp;
				temp=1;
			}
		}
		temp = 1;
		for (int i = 0; i<k; i++) {
			temp*=n;
			temp%=(long)(10e9+7);
		}
//		System.out.println(temp);
//		System.out.println(ptok);
		System.out.println(temp-ptok);
	}
	
	public static long bfs(List<Integer>[] graph, int u, boolean[] visited) {
		int size = 1;
		visited[u] = true;
		Queue<Integer> q = new LinkedList<>();
		q.add(u);
		while (!q.isEmpty()) {
			int cur = q.remove();
			for (Integer v : graph[cur]) {
				if (!visited[v]) {
					visited[v] = true;
					size++;
					q.add(v);
				}
			}
		}
//		System.out.println("Size is "+size);
		return size;
	}
}

class Edge {
	int node1;
	int node2;
	int color;
	
	public Edge(int node1, int node2, int color) {
		this.node1 = node1;
		this.node2 = node2;
		this.color = color;
	}
}
