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 A {

    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer tok;

    public static void main(String[] args) throws IOException {
        try {
            in = new BufferedReader(new FileReader("unequal.in"));
            out = new PrintWriter(new File("unequal.out"));
        } catch (Exception e) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(new OutputStreamWriter(System.out));
        }
        tok = new StringTokenizer("");
        new A().solve();
        out.flush();
        out.close();
    }

    String nextTok() throws IOException {
        while (!tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }

    String nextLine() throws IOException {
        return in.readLine();
    }

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

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

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


    class Pair implements Comparable<Pair> {
        int x;
        int y;
        Pair(int x, int y) {
            this.x = x;
            this.y = y;
        }

        @Override
        public int compareTo(Pair o) {
            if (this.x != o.x) {
                return Integer.compare(this.x, o.x);
            } else {
                return Integer.compare(this.y, o.y);
            }
        }
    }

    int mod = 1000000007;
    ArrayList<Integer>[] g;
    boolean used[];
    int cnt = 0;

    void solve() throws IOException {
        int n = nextInt();
        int k = nextInt();
        g = new ArrayList[n];
        used = new boolean[n];
        for (int i = 0; i < n; i++) {
            g[i] = new ArrayList<>();
        }
        for (int i = 0; i < n - 1; i++) {
            int from = nextInt() - 1;
            int to = nextInt() - 1;
            int color = nextInt();
            if (color == 0) {
                g[from].add(to);
                g[to].add(from);
            }
        }

        long bad = 0;
        for (int i = 0; i < n; i++) {
            if (!used[i]) {
                cnt = 0;
                dfs(i);
                bad = (bad + binPow(cnt, k)) % mod;
            }
        }

        out.println((binPow(n, k) - bad + mod) % mod);

    }

    void dfs(int v) {
        used[v] = true;
        cnt++;
        for (int to: g[v]) {
            if (!used[to]) {
                dfs(to);
            }
        }
    }

    long binPow(long a, long b) {
        long res = 1;
        while (b != 0) {
            if ((b & 1) == 1) {
                res = (res * a) % mod;
            }
            a = (a * a) % mod;
            b >>= 1;
        }
        return res;
    }

}