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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

import static java.lang.System.out;

public class Main {

    private Scanner scanner = new Scanner(System.in);

    private int p[];
    private int size [];

    void make(int v) {
        p[v] = v;
        size[v] = 1;
    }

    private int find(int v) {
        if (v == p[v]) {
            return v;
        }
        return p[v] = find(p[v]);
    }

    private void union(int a, int b) {
        a = find(a);
        b = find(b);
        if (a != b) {
            if (size[a] < size[b]) {
                int t = b;
                b = a;
                a = t;
            }
            p[b] = a;
            size[a] += size[b];
        }
    }


    public static void main(String[] args) {
        new Main().solve();
    }

    private void solve() {
        int n = scanner.nextInt(), k = scanner.nextInt();

        p = new int[n];
        size = new int[n];
        for (int i = 0; i < n; i++) {
            make(i);
        }

        for (int i = 0; i < n - 1; i++) {
            int u = scanner.nextInt() - 1, v = scanner.nextInt() - 1, c = scanner.nextInt();
            if (c == 0) {
                union(u, v);
            }
        }

        long ans = 1;
        for (int i = 0; i < k; i++) {
            ans = (ans * n) % 1000000007;
        }

        boolean used[] = new boolean[n];
        for (int i = 0; i < n; i++) {
            int t = find(i);
            if (! used[t]) {
                int sz = size[t];
                long bad = 1;
                for (int j = 0; j < k; j++) {
                    bad = bad * (sz) % 1000000007;
                }
                ans = ans - bad;
                used[t] = true;
            }
        }
        out.println(ans );
    }
}