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.math.*;
import java.io.*;

public class c {
	class Edge {
		int u, v, c;
		public Edge(int uu, int vv, int cc) {
			u = uu;  v = vv;  c = cc;
		}
	}
	class Tree {
		int n;  List<Edge>[] adj;
		public Tree(int nn) {
			adj = new ArrayList[n = nn];
			for(int i = 0 ; i < n ; adj[i++] = new ArrayList<>());
		}
		void add(int u, int v, int c) {
			adj[u].add(new Edge(u, v, c));
			adj[v].add(new Edge(v, u, c));
		}
		long solve(int k) {
			long all = exp(n, k, MOD);
			long remove = 0;
			boolean[] visited = new boolean[n];
			Deque<Integer> q = new ArrayDeque<>();
			for(int i = 0 ; i < n ; i++) {
				if(!visited[i]) {
					visited[i] = true;
					q.add(i);
					int cmp = 0;
					while(!q.isEmpty()) {
						int u = q.pop();
						cmp++;
						for(Edge e : adj[u]) {
							if(e.c == 0 && !visited[e.v]) {
								visited[e.v] = true;
								q.add(e.v);
							}
						}
					}
					remove += exp(cmp, k, MOD);
					remove %= MOD;
				}
			}
			return (all - remove + MOD) % MOD;
		}
	}
	long exp(long b, long e, long m) {
		if(e == 0) return 1;
		long res = exp(b, e / 2, m);
		res = (res * res) % m;
		if((e & 1) > 0) res = (res * b) % m;
		return res;
	}
	final int MOD = (int) 1e9 + 7;
	public c() {
		FS scan = new FS();
		int n = scan.nextInt(), k = scan.nextInt();
		Tree t = new Tree(n);
		for(int i = 0 ; i < n - 1 ; i++) {
			int u = scan.nextInt() - 1,v = scan.nextInt() - 1, c = scan.nextInt();
			t.add(u, v, c);
		}
		System.out.println(t.solve(k));
	}
	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		public String next() {
			while(!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch(Exception e) { e.printStackTrace(); }
			}
			return st.nextToken();
		}
		public int nextInt() { return Integer.parseInt(next()); }
		public long nextLong() { return Long.parseLong(next()); }
	}
	public static void main(String[] args) { new c(); }
}
