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 javafx.util.*;

public class Master {

    public static Scanner in = new Scanner(System.in);

    public static long pow(long a, long b) {
        long res = 1;
        while (b > 0) {
            if (b % 2 != 0) {
                res *= a;
            }
            b /= 2;
            a *= a;
        }
        return res;
    }

    public static void main(String[] args) throws java.lang.Exception {
        int n = in.nextInt();
        int k = in.nextInt();
        Graph g = new Graph(n);
        int tmp = n - 1;
        while (tmp-- > 0) {
            int u = in.nextInt();
            int v = in.nextInt();
            int type = in.nextInt();
            if (type != 1) {
                g.add(u, v);
            }
        }
        long ans = pow(n, k) % 1000000007;
        ans -= g.solve(k);
        System.out.println(ans);
    }

    public static class Graph {

        int V = 0;
        ArrayList[] data = null;
        int size = 0;

        public Graph(int v) {
            this.V = v + 1;
            data = new ArrayList[this.V];
            for (int i = 0; i <= v; i++) {
                this.data[i] = new ArrayList< Integer>();
            }
        }

        public void add(int u, int v) {

            data[u].add(v);
            data[v].add(u);
        }

        public void dfs(int s, int ID, boolean[] visited) {
            size++;
            visited[s] = true;
            ArrayList<Integer> adj = data[s];
            for (int v : adj) {
                if (!visited[v]) {
                    dfs(v, ID, visited);
                }
            }
        }

        public long solve(int k) {
            int ID = 1;

            int[] compSize = new int[this.V];
            boolean[] v = new boolean[this.V];
            for (int i = 1; i < v.length; i++) {
                if (!v[i]) {
                    dfs(i, ID, v);
                    compSize[ID++] = size;
                    size = 0;
                }
            }
            long ans = 0;

            for (int i = 1; i < ID; i++) {
                int sz = compSize[i];
                ans += pow(sz, k);
            }
            return ans;
        }
    }

}
