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.ArrayList;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class C {
    static BufferedReader br;
    static StringTokenizer st;

    static String nextToken() {
        try {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(br.readLine());
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
        return st.nextToken();
    }

    static String nextString() throws IOException {
        return br.readLine();
    }

    static char nextChar() throws IOException {
        return (char) br.read();
    }

    private static int nextInt() {
        return Integer.parseInt(nextToken());
    }

    static ArrayList<Integer>[] graf;

    public static void main(String[] args) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);
        int n = nextInt();
        int k = nextInt();
        graf = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            graf[i] = new ArrayList<>();
        }
        for (int i = 0; i < n - 1; i++) {
            int x = nextInt() - 1;
            int y = nextInt() - 1;
            int s = nextInt();
            if (s == 0) {
                graf[x].add(y);
                graf[y].add(x);
            }
        }
        tr = new boolean[n];
        for (int i = 0; i < n; i++) {
            if (!tr[i]) {
                b.add(new TreeSet<>());
                id++;
                dfs(i);
            }
        }
        long ans = n;
        for (int i = 1; i < k; i++) {
            ans *= n;
            ans %= 1000000007;
        }
        for (int i = 0; i < b.size(); i++) {
            long km = 1;
            for (int j = 0; j < k; j++) {
                km = km * b.get(i).size() % 1000000007;
            }
            ans -= km;
            ans %= 1000000007;
        }
        System.out.println(ans);
        pw.close();
    }

    static int id = -1;
    static ArrayList<TreeSet<Integer>> b = new ArrayList<>();
    static boolean[] tr;

    private static void dfs(int v) {
        if (tr[v]) {
            return;
        }
        tr[v] = true;
        b.get(id).add(v);
        for (int i = 0; i < graf[v].size(); i++) {
            dfs(graf[v].get(i));
        }
    }
}
