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.IOException;
import java.util.ArrayList;
import java.util.Scanner;
 
public class A
{
    

	static Node[] graph;
    static boolean[] visited;
    static int n;
    static int k;
    static ArrayList<Integer> tab;
    static final int MOD = (int) 1e9 + 7;

    static void dfs(int a) {
        visited[a] = true;
        for (Pair p : graph[a].edges) {
            if (!visited[p.x]) {
                dfs(p.x);
                graph[a].nb += graph[p.x].nb + 1;
            }
        }
    }

    static int pow(long x, long y, int p) {
        long res = 1;
        x = x % p;

        while (y > 0) {
            if ((y & 1) == 1)
                res = (res * x) % p;
            y = y >> 1;
            x = (x * x) % p;
        }
        return (int) res;
    }

    
        


    static class Pair {
        int x;
        int y;

    }

    static class Node {
        ArrayList<Pair> edges;
        int nb;

    }

	
	
	public static void main(String[] args) throws IOException {
		Scanner sc=new Scanner(System.in);
		n = sc.nextInt();
        k = sc.nextInt();
        graph = new Node[n];
        visited = new boolean[n];
        for (int i = 0; i < n; i++) {
            graph[i] = new Node();
            graph[i].edges = new ArrayList<>();
        }
        tab = new ArrayList<>();
        int x, y, t;
        Pair p;
        for (int i = 0; i < n - 1; i++) {
            x = sc.nextInt()- 1;
            y = sc.nextInt() - 1;
            t = sc.nextInt();
            if (t == 0) {
                p = new Pair();
                p.x = y;
                p.y = t;
                graph[x].edges.add(p);
                p = new Pair();
                p.x = x;
                p.y = t;
                graph[y].edges.add(p);
            }
        }
        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                dfs(i);
                tab.add(graph[i].nb + 1);
            }
        }
        long ans = pow(n, k, MOD);
        for (int i = 0; i < tab.size(); i++) {
            ans -= (pow(tab.get(i), k, MOD));
            if (ans < 0) {
                ans += MOD;
            }
        }

        System.out.println(ans);
        

	}

}