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.*; 
import java.util.*; 
  
public class Graph 
{ 
    private int V; 
    private ArrayList<Integer> adj[]; 
  
    Graph(int v) 
    { 
        V = v; 
        adj = new ArrayList[v]; 
        for (int i=0; i<v; ++i) 
            adj[i] = new ArrayList(); 
    } 
    void addEdge(int u,int v) 
    { 
        adj[u].add(v);
        adj[v].add(u);
    }
    static long pow(long l,long k){
        long m=1000000007;
        long power=1;
        for(int i=1;i<=k;i++){
            power=((power%m)*(l%m))%m;
        }
        return power;
    }
    long DFS(int k)
    {
        boolean visited[]=new  boolean[V];
        long sum=0;
        for(int i=0;i<V;i++){
            if(!visited[i]){
                int l=DFSUtil(i,visited);
                sum=sum+pow((long)l,(long)k);
            }
        }
        return sum;
    }
   
  
   int DFSUtil(int s,boolean visited[])
  {
      visited[s]=true;int k=0;
      for (int i=0;i<adj[s].size();i++)
      {
          if(!visited[adj[s].get(i)])
          {
             k=k+DFSUtil(adj[s].get(i),visited);
          }
      }
      return k+1;
  }
  
    public static void main(String args[]) 
    { 
        try{
        Scanner sc=new Scanner(System.in);
        int n=sc.nextInt();
        int k=sc.nextInt();
        Graph g = new Graph(n);
        for(int i=1;i<n;i++)
        {
            int u=sc.nextInt();int v=sc.nextInt();int e=sc.nextInt();
            if(e==0)
            g.addEdge(u-1,v-1);
        }
        long bad=g.DFS(k);
        long ans=pow((long)n,(long)k)-bad;
        System.out.println(ans);
        }catch(Exception e)
        {
            return;
        }
    } 
} 