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.*;

public class Main {
  public static boolean[][] adj;
  public static boolean[] used;

  public static long pow(long n, int k) {
    long result = 1;
    while(k > 0) {
      result *= n;
      result %= 1_000_000_007;
      k--;
    }
    return result;
  }

  public static int count(int i, int n) {
    used[i] = true;
    int sum = 0;
    for(int j = i; j < n; j++) {
      if(adj[i][j]) {
        sum += count(j, n);
      }
    }
    return 1 + sum;
  }

  public static void main(String[] args) {
    Scanner U = new Scanner(System.in);
    
    int n = U.nextInt();
    int k = U.nextInt();

    long sum = pow(n, k);

    adj = new boolean[n][n];
    for(int i = 0; i < n; i++) {
      Arrays.fill(adj[i], false);
    }

    used = new boolean[n];
    Arrays.fill(used, false);

    for(int i = 0; i < n-1; i++) {
      int a = U.nextInt() - 1;
      int b = U.nextInt() - 1;
      boolean isRed;
      if(U.nextInt() == 0){
        isRed = true;
      } else {
        isRed = false;
      }
      
      if(isRed) {
        adj[a][b] = true;
        adj[b][a] = true;
      }
    }

    for(int i = 0; i < n; i++) {
      if(!used[i]) {
        used[i] = true;
        sum -= pow(count(i, n), k);
      }
    }

    System.out.println(sum);

    U.close();
  }
}