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 test {
    
    
       static class DSU {
    
    int[] p;int[] sz;int op;int c;
    public void G(int n)
    {
        p=new int[n];
        sz=new int[n];c=n;
        op=n;
        for(int h=0;h<n;h++)
        {
            sz[h]=1;p[h]=h;
        }
    }
    public int find(int x)
    {
        int y=x;
        while(x!=p[x]) x=p[x];
        while(y!=p[y])
        {
            int tem=p[y];
            p[y]=x;y=tem;
        }
        return p[y];
    }
    public void union(int a,int b)
    {
        int x,y;
        x=find(a);y=find(b);
        if(x==y) return;
        if(sz[x]>sz[y])
        {
            p[y] = x;
            sz[x]+=sz[y];
        }
        else
        {
            p[x]=y;sz[y]+=sz[x];
        }
        c--;
        
    }}
    static long mod=1000000000+7l;
    static long power(long x, long y)
	  {
	    long res = 1l; // Initialize result
		long p=mod;
	    x = x % p; // Update x if it is more than or
	    // equal to p
	 
	    if (x == 0l)
	      return 0; // In case x is divisible by p;
	 
	    while (y > 0)
	    {
	 
	      // If y is odd, multiply x with result
	      if ((y & 1) != 0)
	        res = (res * x) % p;
	 
	      // y must be even now
	      y = y >> 1; // y = y/2
	      x = (x * x) % p;
	    }
	    return res;
	  }
    public static void main (String[] args) throws Exception {
        
        BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
        
        int y=1;//Integer.parseInt(bf.readLine());
        for(int h=0;h<y;h++)
        {
           String[] xl=(bf.readLine()).split(" ");
            //long a,b,c,d;long ans=0l;
            int n,t;
            n=Integer.parseInt(xl[0]);
            t=Integer.parseInt(xl[1]);
            
            DSU ds=new DSU();
            ds.G(n);
           for(int o=0;o<n-1;o++)
           {
           	  String[] lk = bf.readLine().split(" ");
           	  int a = Integer.parseInt(lk[0])-1;
           	  int b=Integer.parseInt(lk[1])-1;
           	  int c=Integer.parseInt(lk[2]);
           	  if(c==0) ds.union(a,b);
           }
           long sum=0l;
           int[] arr=ds.sz;
           
           for(int u=0;u<arr.length;u++)
           {
           	   if(ds.find(u) == u)
           	   {
           	   	   sum = (sum%mod + power(arr[u]+0l,t+0l)%mod)%mod;
           	   }
           }
           sum = power(n+0l,t+0l) - sum;
           if(sum<0) sum= (sum+mod)%mod;
           System.out.println(sum);
            
        }
    }
}
/*

2
1000000000 1000000000
aaabb

*/


