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 ProblemC {

    static int MOD = 1000000007;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
//        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);

        int nodes = sc.nextInt();
        int len = sc.nextInt();
        DisjointSet ds = new DisjointSet(nodes);
        for (int i = 0; i < nodes-1; i++) {
            int ori = sc.nextInt();
            int des = sc.nextInt();
            int is = sc.nextInt();
            if (is == 0){
                ds.join(ori,des);
            }
        }
        long sol = pow(nodes,len,MOD);
        for (int j = 1; j <= nodes ; j++) {
            int compSize = ds.si[ds.father(j)];
            sol = (sol - pow(compSize,len-1,MOD) + MOD)%MOD;
        }

        pw.println(sol);


        pw.flush();
    }

    static long pow(long base, long exp, long mod) {
        long res = 1;
        base = base % mod;
        while (exp > 0) {
            if((exp & 1)==1)
                res = (res * base) % mod;
            exp = exp/2;
            base = (base * base) % mod;
        }
        return res;
    }

    static class DisjointSet{
        public int[] fa;
        public int[] si;
        public int components;

        public DisjointSet(int size) {
            this.fa = new int[size+4];
            this.si = new int[size+4];
            this.components = size;
            for (int i = 0; i < size + 4; i++) {
                this.fa[i] = i;
                this.si[i] = 1;
            }
        }

        public int father (int node){
            while (fa[node]!=node){
                node = fa[node];
            }
            return node;
        }

        public boolean join(int n1, int n2){
            int fa1 = father(n1);
            int fa2 = father(n2);
            if (fa1==fa2)return false;
            components--;
            if (si[fa1]>si[fa2]){
                fa[fa2]=fa1;
                si[fa1]+=si[fa2];
            } else {
                fa[fa1]=fa2;
                si[fa2]+=si[fa1];
            }
            return true;
        }

    }
}
