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.util.*;

public class Main {
    static final int MOD = 1000000007;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int k = sc.nextInt();
        int[] u = new int[n - 1];
        int[] v = new int[n - 1];
        int[] x = new int[n - 1];
        for (int i = 0; i < n - 1; i++) {
            u[i] = sc.nextInt();
            v[i] = sc.nextInt();
            x[i] = sc.nextInt();
        }
        System.out.println(solve(u, v, x, k));

        sc.close();
    }

    static int solve(int[] u, int[] v, int[] x, int k) {
        int n = u.length + 1;

        int[] parents = new int[n];
        Arrays.fill(parents, -1);

        for (int i = 0; i < x.length; i++) {
            if (x[i] == 0) {
                int root1 = findRoot(parents, u[i] - 1);
                int root2 = findRoot(parents, v[i] - 1);

                if (root1 != root2) {
                    parents[root2] = root1;
                }
            }
        }

        Map<Integer, Integer> rootToCount = new HashMap<>();
        for (int i = 0; i < n; i++) {
            int root = findRoot(parents, i);

            rootToCount.put(root, rootToCount.getOrDefault(root, 0) + 1);
        }

        int result = powMod(n, k);
        for (int count : rootToCount.values()) {
            result = subtractMod(result, powMod(count, k));
        }

        return result;
    }

    static int findRoot(int[] parents, int node) {
        int root = node;
        while (parents[root] != -1) {
            root = parents[root];
        }

        int p = node;
        while (p != root) {
            int next = parents[p];
            parents[p] = root;

            p = next;
        }

        return root;
    }

    static int powMod(int base, int exponent) {
        int result = 1;
        for (int i = 0; i < exponent; i++) {
            result = multiplyMod(result, base);
        }

        return result;
    }

    static int subtractMod(int x, int y) {
        return (x - y + MOD) % MOD;
    }

    static int multiplyMod(int x, int y) {
        return (int) ((long) x * y % MOD);
    }
}