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.Scanner;

public class edgytrees {
    static final int MOD = (int) (Math.pow(10, 9) + 7);
    public static void main(String[] args) {
       Scanner in = new Scanner(System.in);
       int N = in.nextInt();
       int k = in.nextInt();
       long total = modExponentiate(N, k);
       DisjointUnionSets set = new DisjointUnionSets(N);
       for (int i = 0; i < N-1; i++) {
           int a = in.nextInt()-1;
           int b = in.nextInt() - 1;
           int decider = in.nextInt();
           if (decider == 0) {
               set.union(a, b);
           }
       }
       for (int i = 0; i < N; i++) {
           if (set.parent[i] == i) {
               total -= modExponentiate(set.rank[i], k);
           }
       }
       while (total < 0) {
           total += MOD;
       }
       System.out.println(total);
    }
    public static long modExponentiate(int base, int power) {
        long val = 1;
        while (power > 0) {
            val *= base;
            val %= MOD;
            power--;
        }
        return val;
    }
    static class DisjointUnionSets {
        int[] rank, parent;
        int n;
        int size;

        // Constructor
        public DisjointUnionSets(int n) {
            size = n;
            rank = new int[n];
            parent = new int[n];
            this.n = n;
            makeSet();
        }

        // Creates n sets with single item in each
        void makeSet() {
            for (int i = 0; i < n; i++) {
                // Initially, all elements are in
                // their own set.
                parent[i] = i;
                rank[i] = 1;
            }
        }

        // Returns representative of x's set
        int find(int x) {
            // Finds the representative of the set
            // that x is an element of
            if (parent[x] != x) {
                // if x is not the parent of itself
                // Then x is not the representative of
                // his set,
                parent[x] = find(parent[x]);

                // so we recursively call Find on its parent
                // and move i's node directly under the
                // representative of this set
            }

            return parent[x];
        }

        // Unites the set that includes x and the set
        // that includes x
        void union(int x, int y) {
            // Find representatives of two sets
            int xRoot = find(x), yRoot = find(y);

            // Elements are in the same set, no need
            // to unite anything.
            if (xRoot == yRoot)
                return;
            rank[xRoot] = rank[xRoot]+rank[yRoot];

            // If x's rank is less than y's rank
            if (rank[xRoot] < rank[yRoot])

                // Then move x under y  so that depth
                // of tree remains less
                parent[xRoot] = yRoot;

                // Else if y's rank is less than x's rank
            else if (rank[yRoot] < rank[xRoot])

                // Then move y under x so that depth of
                // tree remains less
                parent[yRoot] = xRoot;

            else // if ranks are the same
            {
                // Then move y under x (doesn't matter
                // which one goes where)
                parent[yRoot] = xRoot;

                // And increment the the result tree's
                // rank by 1
            }
            size--;

        }
    }
}
