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.*;

public class Edgy_Tress
{
    static ArrayList<Integer> arr [];
    static ArrayList<Integer> visited;
    static long mod = 1000000007;
    static long DFS(int start,int n)
    {
        visited.set(start,1);
        //System.out.print(start+" ");
        Iterator<Integer> i = arr[start].listIterator();
        long count = 1;
        while(i.hasNext())
        {
            int next = i.next();
            //System.out.print(next+" ");
            if(visited.get(next)==0)
                count+=DFS(next,n);
        }
        return count;


    }
    static long pow(long a, int b)
    {
        if(b==0)
            return 1;
        long ans = pow(a,b/2);
        if(b%2==0)
            return ans*ans%mod;
        else
            return ans*ans%mod*a%mod;
    }
    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int  n =sc.nextInt();
        int  k =sc.nextInt();
        arr=  new ArrayList[n+1];
        visited = new ArrayList<>();

        for(int i =0;i<=n;i++)
        {
            arr[i] = new ArrayList<>();

            visited.add(0);

        }

        for(int i=0;i<n-1;i++)
        {
            int v1 =sc.nextInt(), v2 = sc.nextInt();
            int c = sc.nextInt();
            if(c==0)
            {
                arr[v1].add(v2);
                arr[v2].add(v1);
            }

        }
        long bad=0;

        for(int i=1;i<=n;i++)
        {
            if(visited.get(i)==0)
            {
                bad+= (long)(pow(DFS(i,n),k));

                //System.out.println(bad);
            }
        }
        //System.out.println((long)Math.pow(n,k)-bad);
        bad%=mod;
        long ans = (pow(n,k)-bad+mod)%mod;
        System.out.print(ans);




    }
}
