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 java.math.*;
public class Main{
	static int n,k,a,b,c,dex;
	static int rank[]=new int[100005],num[]=new int[100005],pre[]=new int[100005];
	static void init(){
		for(int i=1;i<=n;i++){
			rank[i]=1;
			pre[i]=i;
		}
	}
	static int find_pre(int x){
		if(pre[x]==x) return x;
		return pre[x]=find_pre(pre[x]);
	}
	static void union(int x,int y){
		int px=find_pre(x);
		int py=find_pre(y);
		if(px==py) return;
		if(rank[px]<rank[py]){
			pre[px]=py;
		}
		else{
			if(rank[px]==rank[py]){
				rank[px]++;
			}
			pre[py]=px;
		}
	}
	public static void main(String argas[]){
		Scanner cin=new Scanner(System.in);
		dex=1;
		n=cin.nextInt();
		k=cin.nextInt();
		init();
		for(int i=2;i<=n;i++){
			a=cin.nextInt();
			b=cin.nextInt();
			c=cin.nextInt();
			if(c==0){
				union(a,b);
			}
		}
		for(int i=1;i<=n;i++){
			num[pre[i]]++;
		}
		long ans=1,uu=(long)(1e9+7),tt=n,temp;
        for(int i=1;i<=k;i++){
			ans=(ans*tt)%uu;
		}
        ans-=n;
        if(ans<0) ans+=uu;
		for(int i=1;i<=n;i++){
			if(num[i]>0){
				temp=1;
				tt=num[i];
				for(int j=1;j<=k;j++){
					temp=(temp*tt)%uu;
				}
				temp-=num[i];
				if(temp<0) temp+=uu;
				ans-=temp;
				if(ans<0) ans+=uu;
			}
		}
		System.out.println(ans);
	}
}