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 {

   public static final int MOD = (int) 1e9 + 7;
   public static void main(String[] args) {
      Scanner sc = new Scanner(System.in);
      
      int n = sc.nextInt();
      int k = sc.nextInt();
      
      Node[] graph = new Node[n];
      for(int i = 0; i < n; i++) {
         graph[i] = new Node();
         graph[i].val = i;
      }
      
      for(int i = 0; i < n - 1; i++) {
         int a = sc.nextInt() - 1;
         int b = sc.nextInt() - 1;
         int color = sc.nextInt();
         
         Edge uv = new Edge();
         uv.u = a;
         uv.v = b;
         uv.c = color;
         graph[a].edges.add(uv);
         
         Edge vu = new Edge();
         vu.u = b;
         vu.v = a;
         vu.c = color;
         graph[b].edges.add(vu);
      }
      
      //math.pow(n,k) % mod
      long ans = n;
      for(int i = 0; i < k - 1; i++) {
         ans *= n;
         ans %= MOD;
      }
      
      //subtract off [x, x, x]...
      ans -= n;
      
      boolean[] vis = new boolean[n];
      ArrayList<Integer> comp = new ArrayList<Integer>();
      //group components by edge color
      for(int i = 0; i < n; i++) {
         if(!vis[i]) {
            ArrayList<Node> red = new ArrayList<Node>();
            Stack<Node> stack = new Stack<Node>();
            stack.push(graph[i]);
            while(!stack.isEmpty()) {
               Node cur = stack.pop();
               vis[cur.val] = true;
               for(Edge e : cur.edges) {
                  if(e.c == 0 && !vis[e.v]) {
                     red.add(cur);
                     stack.push(graph[e.v]);
                  }
               }
            }
            comp.add(red.size());
         }
      }
      
      for(int i = 0; i < comp.size(); i++) {
         ans -= Math.pow(comp.get(i) + 1, k) % MOD;
         ans += comp.get(i) + 1;
         ans %= MOD;
      }
      System.out.println(ans);
   }
}
class Node {
   public int val;
   public ArrayList<Edge> edges = new ArrayList<Edge>();
}
class Edge {
   public int u;
   public int v;
   public int c;
}