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 C {
    class Pair {
        int to;
        int c;

        public Pair(int to, int c) {
            this.to = to;
            this.c = c;
        }
    }

    int mod = 1000000007;

    long pow(long a, int n) {
        long res = 1;
        while (n > 0) {
            if (n % 2 == 1)
                res = (res * a) % mod;
            a = (a * a) % mod;
            n >>= 1;
        }
        return res;
    }

    public void solve() throws IOException {
        int n = nextInt();
        int k = nextInt();
        ArrayList<Pair>[] g = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            g[i] = new ArrayList<>();
        }
        for (int i = 0; i < n - 1; i++) {
            int x = nextInt() - 1;
            int y = nextInt() - 1;
            int c = nextInt();
            Pair v1 = new Pair(y, c);
            Pair v2 = new Pair(x, c);
            if (c == 0) {
                g[x].add(v1);
                g[y].add(v2);
            }
        }
        ArrayList<Integer> ks = new ArrayList<>();
        boolean[] used = new boolean[n];
        Stack<Integer> dfs = new Stack<>();
        for (int i = 0; i < n; i++) {
            if (!used[i]) {
                boolean hasRed = false;
                for (int j = 0; j < g[i].size(); j++) {
                    if (g[i].get(j).c == 0) {
                        hasRed = true;
                    }
                }
                if (hasRed) {
                    int count = 1;
                    used[i] = true;
                    dfs.add(i);
                    while (!dfs.isEmpty()) {
                        int cur = dfs.pop();
                        for (int j = 0; j < g[cur].size(); j++) {
                            int t = g[cur].get(j).to;
                            if (!used[t]) {
                                dfs.add(t);
                                used[t] = true;
                                count++;
                            }
                        }
                    }
                    ks.add(count);
                }
            }
        }
        long ans = pow(n, k);
        for (int i = 0; i < ks.size(); i++) {
            ans = (ans - pow(ks.get(i), k) + ks.get(i)) % mod;
        }
        System.out.println(Math.max(0, ans - n));
    }


    public void run() {
        try {
            br = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    BufferedReader br;
    StringTokenizer in;
    PrintWriter out;

    public String nextToken() throws IOException {
        while (in == null || !in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    public int[] nextArr(int n) throws IOException {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextInt();
        }
        return res;
    }

    public static void main(String[] args) throws IOException {
        Locale.setDefault(Locale.US);
        new C().run();
    }
}
