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 File {
	static class DSU {
		int[] parents;
		int[] size;
		
		public DSU(int N) {
			parents = new int[N];
			for (int i = 0; i < N; i++) {
				parents[i] = i;
			}
			
			size = new int[N];
		}
		
		public int getParent(int x) {
			if (x != parents[x]) parents[x] = getParent(parents[x]);
			return parents[x];
		}
		
		public void union(int x, int y) {
			int xParent = getParent(x);
			int yParent = getParent(y);
			
			if (xParent != yParent) {
				if (size[xParent] < size[yParent]) {
					parents[xParent] = yParent;
				}
				else if (size[xParent] > size[yParent]) {
					parents[yParent] = xParent;
				}
				else {
					parents[yParent] = xParent;
					size[xParent]++;
				}
			}
		}
	}
	
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
				
		long n = sc.nextInt(); // Size of tree.
		long k = sc.nextInt(); // Length of vertex sequence.
		
		long mod = 1000000007;
		
		boolean hasBlackEdge = false;
		
		DSU dsu = new DSU((int)n);
		
		for (int i = 0; i < n-1; i++) {
			int u = sc.nextInt() - 1;
			int v = sc.nextInt() - 1;
			int color = sc.nextInt(); // 0=red, 1=black
			
			if (color == 1) {
				hasBlackEdge = true;
			}
			else {
				dsu.union(u, v);
			}
		}
		
		// There are no black edges in this graph, so no valid paths.
		if (!hasBlackEdge) {
			System.out.println(0);
			return;
		}
		
		long totalSequences = fastPow(n, k, mod) % mod;
		
		// Count how many nodes have a parent in the DSU.
		Map<Integer, Integer> sizes = new HashMap<>();
		for (int i = 0; i < n; i++) {
			int parent = dsu.getParent(i);
			sizes.put(parent, sizes.getOrDefault(parent, 0) + 1);
		}
		
		long invalidSequences = 0;
		for (int parent : sizes.keySet()) {
			invalidSequences = (invalidSequences + fastPow(sizes.get(parent), k, mod)) % mod;
		}
		
		if (totalSequences < invalidSequences) {
		    totalSequences = (totalSequences + mod);
		}
		
		// Count the total invalid sequences.
		long res = (totalSequences - invalidSequences) % mod;
		
		System.out.println(res);
	}
	
	public static long fastPow(long x, long pow, long mod) {
		if (pow == 0) return 1;
		if (pow == 1) return x;
		
		long half = fastPow(x, pow / 2, mod) % mod;
		
		if (pow % 2 == 0) {
			return (half * half) % mod;
		}
		else {
			return (((half * half) % mod) * x) % mod;
		}
	}
}