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 main {
	static int parent[], rank[];
	static int find(int x) {
		if(parent[x] != x)
			parent[x] = find(parent[x]);
		return parent[x];
	}
	static void union(int a, int b) {
		int x = find(a), y = find(b);
		if(x == y)
			return;
		if(rank[x] > rank[y])
			parent[y] = x;
		else if(rank[x] < rank[y])
			parent[x] = y;
		else {
			parent[y] = x;
			rank[x]++;
		}
	}
	public static final int MOD = 1000000000 + 7;
	static long fastExp(long x, long y) { 
        long Ans = 1L;      
        while(y > 0) { 
            if((y & 1) == 1) 
                Ans = (Ans * x) % MOD; 
            y = y >> 1;  
            x = (x * x) % MOD;  
        } 
        return Ans % MOD; 
    }
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		String s[] = br.readLine().split(" ");
		int N = Integer.parseInt(s[0]), k = Integer.parseInt(s[1]);
		parent = new int[N];
		rank = new int[N];
		for(int i = 0 ; i < N ; i++)
			parent[i] = i;
		for(int i = 0 ; i < N - 1 ; i++) {
			s = br.readLine().split(" ");
			if(s[2].equals("1"))
				continue;
			union(Integer.parseInt(s[0]) - 1 ,Integer.parseInt(s[1]) - 1);
		}
		HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
		for(int i = 0 ; i < N ; i++) {
			int key = find(parent[i]);
			if(hm.containsKey(key))
				hm.put(key, hm.get(key) + 1);
			else
				hm.put(key, 1);
		}
		long Ans = fastExp(N, k);
		for(Map.Entry<Integer, Integer> entry : hm.entrySet()) {
			Ans -= fastExp(entry.getValue(), k);
			Ans = (Ans + MOD) % MOD;
		}
		pw.println(Ans);
		pw.flush();
	}
}