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.
//package albert.codeforces.edgytrees1139c;

import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Solution {
    static long MOD = (long) (1e9 + 7);
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int k = scanner.nextInt();
        int[] uf = new int[n];
        for (int i = 1; i < n; i++) {
            uf[i] = i;
        }
        for (int i = 0; i < n - 1; i++) {
            int s = scanner.nextInt() - 1;
            int d = scanner.nextInt() - 1;
            int c = scanner.nextInt();
            if (c == 0) {
                int sp = parent(s, uf);
                int dp = parent(d, uf);
                uf[sp] = dp;
            }
        }
        Map<Integer, Integer> componentSize = new HashMap<>();
        for (int i = 0; i < n; i++) {
            int p = parent(i, uf);
            componentSize.put(p, componentSize.getOrDefault(p, 0) + 1);
        }
        long ans = 0;

        for (Integer v : componentSize.values()) {
            ans += npowk(v, k);
        }
        System.out.println((npowk(n, k) + MOD - ans) % MOD);
    }

    private static long npowk(int n, int k) {
        long ans = 1;
        for (int i = 0; i < k; i++) {
            ans = ans * n;
            ans = ans % MOD;
        }
        return ans;
    }

    private static int parent(int n, int[] uf) {
        int parent = n;
        while (uf[parent] != parent) {
            parent = uf[parent];
        }
        uf[n] = parent;
        return parent;
    }
}
