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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Solution {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();


        List<List<Integer>> graph = new ArrayList<>();
        for(int i = 0; i<n; i++) {
            graph.add(new ArrayList<>());
        }

        for(int i = 0; i<n-1; i++) {
            int i1 = sc.nextInt();
            int i2 = sc.nextInt();
            int c = sc.nextInt();
            if(c == 0) {
                graph.get(i1-1).add(i2-1);
                graph.get(i2-1).add(i1-1);
            }
        }

        boolean[] visited = new boolean[n];
        for(int i = 0; i<n; i++) {
            visited[i] = false;
        }

        Long tot_count = power(n, k, 1000000007);

        for(int i = 0; i<n; i++) {
            int count = 0;
            if(visited[i] == false) {
                count = dfs(graph, visited, i);
            }
            tot_count -= power(count, k, 1000000007);
            if(tot_count < 0) {
                tot_count += 1000000007;
            }
        }
        System.out.println(tot_count % 1000000007);

    }

    private static int dfs(List<List<Integer>> graph, boolean[] visited, int i) {

        int count = 1;
        visited[i] = true;

        for(int node : graph.get(i)) {
            if(visited[node] == false) {
                visited[node] = true;
                count += dfs(graph, visited, node);
            }
        }

        return count;
    }

    static Long power(int x,  int y, int p)
    {
        System.out.println(x + " " + y);
        Long res = 1L;      // Initialize result
        x = x % p;  // Update x if it is more than or
        // equal to p

        while (y > 0)
        {
            res = (res*x) % p;
            if(res < 0) {
                System.out.println(x + " " + y  + " " + res);
            }
            // y must be even now
            y = y-1; // y = y/2
        }
        return res;
    }
}
