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 static java.lang.Math.*;

public class SolutionB {
       public static final long BASE = 100001;
       public static final int MOD = 1000000007;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		Map<Long, Integer> colors = new HashMap<Long, Integer>();
		Map<Integer, List<Integer>> tree = new HashMap<Integer, List<Integer>>();
		for(int m = 0; m < n-1; m++){
		       int u = sc.nextInt();
		       int v = sc.nextInt();
		       int c = sc.nextInt();
		       if(!tree.containsKey(u))tree.put(u, new ArrayList<Integer>());
		       tree.get(u).add(v);
		       if(!tree.containsKey(v))tree.put(v, new ArrayList<Integer>());
		       tree.get(v).add(u);
		       long key = (long)u + (long)v * BASE;
		       colors.put(key, c);
		       key = (long)v + (long)u * BASE;
		       colors.put(key, c);
		}
		long ans = n;
		for(int i = 0; i < k-1; i++){
		       ans *= n;
		       ans %= MOD;
		}
		ans -= n;
		//System.out.println("ans " + ans);
		boolean visited[] = new boolean[n + 1];
		for(int i = 1; i <= n; i++){
		       if(visited[i])continue;
		       Queue<Integer> bfs = new LinkedList<Integer>();
		       visited[i] = true;
		       bfs.offer(i);
		       int amnt = 0;
		       while(!bfs.isEmpty()){
		              int father = bfs.remove();
		              amnt++;
		              for(int child : tree.get(father)){
		                     if(visited[child])continue;
		                     long key = (long)father + (long)child * BASE;
		                     if(colors.get(key) == 0){
		                            visited[child] = true;
		                            bfs.offer(child);
		                     }
		              }
		       }
		       //System.out.println( "amount " + amnt);
		       if(amnt < 2)continue;
		       long comb = amnt;
		       for(int j = 0; j < k - 1; j++){
		              comb *= amnt;
		              comb %= MOD;
		       }
		       comb -= amnt;
		       ans -= comb;
		}
		System.out.println(ans);
	}
}