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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.AbstractCollection;
import java.util.Iterator;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        int mod = 1000000009;
        long combin = 0;
        boolean[] visited;
        ArrayList<ArrayList<Integer>> graph = new ArrayList<>();

        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int k = in.nextInt();
            combin = power(n, k);
            visited = new boolean[n + 1];
            for (int i = 0; i <= n; i++) {
                graph.add(new ArrayList<>());
            }
            for (int i = 0; i < n - 1; i++) {
                int u = in.nextInt();
                int v = in.nextInt();
                int x = in.nextInt();
                if (x == 0) {
                    graph.get(u).add(v);
                    graph.get(v).add(u);
                }
            }
            for (int i = 1; i <= n; i++) {
                if (!visited[i]) {
                    int count = 0;
                    PriorityQueue<Integer> q = new PriorityQueue<>();
                    q.add(i);
                    while (!q.isEmpty()) {
                        int p = q.poll();
                        if (!visited[p]) {
                            count++;
                            visited[p] = true;
                            Iterator<Integer> x = graph.get(p).iterator();
                            while (x.hasNext()) {
                                int m = x.next();
                                if (!visited[m]) q.add(m);
                            }
                        }
                    }

                    combin = combin % mod - power(count, k) % mod;
                    combin = combin % mod;
                }
            }
            out.println(combin);
        }

        long power(long a, long b) {
            if (b == 0) return 1;
            return ((a % mod) * power(a, b - 1 % mod) % mod);
        }

    }
}

