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.PrintWriter;
import java.util.ArrayList;
import java.util.Random;
import java.util.Scanner;
import java.util.TreeSet;

public class Main {
    static Scanner in = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out);
    static int parent[];
    static int size[];
    static long MOD = (long) 1e9 + 7;
    static Random rand = new Random();

    static void makeSet (int u) {
        parent[u] = u;
        size[u] = 1;
    }

    static int findSet (int u) {
        if (u == parent[u]) return u;
        parent[u] = findSet(parent[u]);
        return parent[u];
    }

    static void unionSet (int u, int v) {
        int parentU = findSet(u);
        int parentV = findSet(v);
        if (parentU != parentV) {
            if (size[parentU] < size[parentV]) {
                parent[parentU] = parentV;
                size[parentV] += size[parentU];
            } else {
                parent[parentV] = parentU;
                size[parentU] += size[parentV];
            }
        }
    }

    static long fast_pow (long base, long exponent) {
        if (exponent == 0) {
            return 1;
        }
        long mid = fast_pow(base, exponent / 2);
        if (exponent % 2 == 0) {
            return mid * mid % MOD;
        } else {
            return mid * mid % MOD * base % MOD;
        }
    }

    public static void main(String[] args) {
        int n = in.nextInt();
        int k = in.nextInt();

        parent = new int[n];
        size = new int[n];
        for (int i = 0; i < n; i++) {
            makeSet(i);
        }

        for (int i = 0; i < n - 1; i++) {
            int u = in.nextInt() - 1;
            int v = in.nextInt() - 1;
            int color = in.nextInt();
            if (color == 0) {
                unionSet(u, v);
            }
        }

        // get unique parent
        TreeSet<Integer> uniqueParent = new TreeSet<>();
        for (int i = 0; i < n; i++) {
            uniqueParent.add(findSet(i));
        }

        Object uniqueParentList[] = uniqueParent.toArray();
        // count inverse graph
        long inverseGraph = 0;
        for (Object uParent : uniqueParentList) {
            inverseGraph = (inverseGraph + fast_pow(size[(Integer)uParent], k)) % MOD;
            // out.println("unique Parent: " + ((Integer)uParent + 1));
        }

        // for (int i = 0; i < n; i++) {
        //     out.println("i: " + (i + 1) + " parent: " + (findSet(i) + 1) + " size: " + size[i]);
        // }

        long sum = fast_pow(n, k);
        sum = (sum - inverseGraph + MOD) % MOD;

        out.println(sum);

        in.close();
        out.close();
    }
}