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.
/*
Roses are red
Memes are neat
All my test cases time out
Lmao yeet
*/
import java.util.*;
import java.io.*;

   public class x1139C
   {
      public static void main(String args[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int N = Integer.parseInt(st.nextToken());
         int K = Integer.parseInt(st.nextToken());
         LinkedList<Integer>[] edges = new LinkedList[N+1];
         for(int i=0; i <= N; i++)
            edges[i] = new LinkedList<Integer>();
         //floof
         for(int i=0; i < N-1; i++)
         {
            st = new StringTokenizer(infile.readLine());
            int a = Integer.parseInt(st.nextToken());
            int b = Integer.parseInt(st.nextToken());
            int c = Integer.parseInt(st.nextToken());
            if(c == 0)
            {
               edges[a].add(b);
               edges[b].add(a);
            }
         }
         final long MOD = 1000000007L;
         long res = power(N, (long)K, MOD);
         //PIE
         boolean[] seen = new boolean[N+1];
         for(int i=1; i <= N; i++)
         {
            long ret = 0L;
            if(!seen[i] && edges[i].size() > 0)
            {
               Queue<Integer> q = new LinkedList<Integer>();
               q.add(i);
               seen[i] = true;
               while(q.size() > 0)
               {
                  int curr = q.poll();
                  ret++;
                  for(int next: edges[i])
                     if(!seen[next])
                     {
                        q.add(next);
                        seen[next] = true;
                     }
               }
            }
            else if(!seen[i])
               ret++;
            res -= power(ret, (long)K, MOD);
            while(res < 0) 
               res += MOD;
         }
         System.out.println(res);
      }
      static long power(long x, long y, long p) 
       { 
           long res = 1L;      
           x = x % p;  
         
           while (y > 0) 
           { 
               if((y & 1)==1) 
                   res = (res * x) % p; 
               y = y >> 1;  
               x = (x * x) % p;  
           } 
           return res; 
       } 
   }