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 {

    static long mod = (long)1e9 + 7;
    static ArrayList<Pair>[] gr;
    static boolean[] used;

    public static void main(String[] args) {

        Scanner in = new Scanner(System.in);
        int n = in.nextInt(), k = in.nextInt();
        gr = new ArrayList[n];
        for(int i = 0; i < n; i++) {
            gr[i] = new ArrayList<>();
        }
        used = new boolean[n];
        for(int i = 0; i < n - 1; i++) {
            int u = in.nextInt() - 1, v = in.nextInt() - 1, x = in.nextInt();
            gr[u].add(new Pair(v, x));
            gr[v].add(new Pair(u, x));
        }

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

        for(int i = 0; i < n; i++) {
            if(! used[i]) {
                used[i] = true;
                long count = dfs(i) + 1;
                long cur = 1;
                for(int j = 0; j < k; j++) {
                    cur = (cur * count) % mod;
                }
                //println(ans + " " + count + " " + cur);
                ans = (ans - cur) % mod;
            }
        }
        println(ans);
    }

    static long dfs(int v) {
        int count = 0;
        for(Pair p : gr[v]) {
            if(! used[p.a] && p.b != 1) {
                used[p.a] = true;
                count++;
                count += dfs(p.a);
            }
        }
        return count;
    }

    static class Pair {
        int a, b;
        Pair(int a, int b) {
            this.a = a;
            this.b = b;
        }
    }

    static void print(Object o) { System.out.print(o.toString()); }

    static void println(Object o) {
        System.out.println(o.toString());
    }
}
