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.
//22 march
#include<bits/stdc++.h>
using namespace std;
vector<int>adj[100005];
int parent[100005];
int rank1[100005];
#define M 1000000007
#define ll long long
int find(int a){
if(parent[a]==a)return a;
return parent[a]=find(parent[a]);
}
ll power(ll a,ll b)
{
	if(b==0)
	return 1;
	if(b%2==0)
	return power((a*a)%M,b/2);
	else
	return a*power((a*a)%M,(b-1)/2)%M;
}
void merge(int a,int b)
{
	a=find(a);
	b=find(b);
	if(a!=b)
	{
		if(rank1[a]<rank1[b])
		{
			swap(a,b);
	    }
	    parent[b]=a;
	    for(int x:adj[b])
	    adj[a].push_back(x);
	    rank1[a]+=rank1[b];
	    
	}
}
int main(){
int n,k;
cin>>n>>k;
for(int i=1;i<=n;i++)
{parent[i]=i;adj[i].push_back(i);rank1[i]=0;}
int ind=-1;
for(int i=0;i<n-1;i++)
{
	int u,v,x;
	cin>>u>>v>>x;
	if(x==0)
	{
		merge(u,v);
		ind=0;
    }
}
if(ind!=-1){
int flg[n+1]={0};
ll cnt,sub=0,p=0;
for(int i=1;i<=n;i++)
{   int f=find(i);
	//cout<<adj[i].size()<<endl;
	if(flg[f]==0 && adj[f].size()!=1)
	{cnt=adj[f].size();flg[f]=1;sub+=power(cnt,k);p+=cnt;}
}
//cout<<sub<<endl;
ll ans=power(n,k);
ans= ans-sub-(n-p);
cout<<ans%M;
}
else{
ll ans=power(n,k)-n;

cout<<ans%M;
}
}
