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 EdgyTree {

    static int mod = 1000000007;

    public static void main(String[] args) {

        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int k = scan.nextInt();

        Graph graph = new Graph(n);

        boolean[] visited = new boolean[n];
        long total = power(n, k, mod);

        for(int i = 0; i < n - 1; i++) {

            int u = scan.nextInt()-1;
            int v = scan.nextInt()-1;
            int x = scan.nextInt();

            if(x == 0){
                graph.union(u, v);
            }
        }

        for(int i = 0; i < n; i++) {

            int p = graph.findParent(i);

            if(visited[p]){
                continue;
            }

            visited[p] = true;
            long now = power(graph.len[p], k, mod);
            total = (total - now + mod) % mod;
        }

        System.out.println(total);
    }

    static long power(long x, long y, int p) {
        long res = 1;
        x = x % p;

        while (y > 0) {
            if ((y & 1) == 1) {
                res = (((res % p) * (x % p)) % p);
            }

            y = y >> 1;
            x = (((x % p) * (x % p)) % p);
        }
        return res;
    }

}

class Graph {

    int[] root;
    int[] len;
    int n;

    Graph(int n) {
        this.n = n;
        root = new int[n];
        len = new int[n];
        for (int i = 0; i < n; i++) {
            root[i] = i;
            len[i] = 1;;
        }
    }

    int findParent(int i) {
        while(root[i]!=i) {
            root[i] = root[root[i]];
            i=root[i];
        }
        return i;
    }

    boolean union(int x, int y) {

        int parX = findParent(x);
        int parY = findParent(y);

        if (parX == parY)
            return false;

        if (len[parX] < len[parY]) {
            root[parY] = parX;
            len[parX] += len[parY];
        } else {
            root[parX] = parY;
            len[parY] += len[parX];
        }

        return true;
    }
}