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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class EdgyTrees {
    static class Edge {
        int to;
        int col;

        public Edge(int to, int col) {
            this.to = to;
            this.col = col;
        }
    }

    static int res = 0;

    static void dfs(List<List<Edge>> graph, boolean[] visited, int cur) {
        visited[cur] = true;
        res++;
        for (Edge e : graph.get(cur)) {
            if (e.col == 1 || visited[e.to]) continue;
            dfs(graph, visited, e.to);
        }
    }

    static final long mod = 1_000_000_000 + 7;

    static long pow(long base, long exp) {
        long ans = 1;
        while (exp > 0) {
            if (exp % 2 == 0) {
                exp /= 2;
                base = base * base;
                base %= mod;
            } else {
                exp--;
                ans *= base;
                ans %= mod;
            }
        }
        return ans % mod;
    }

    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer stringTokenizer = new StringTokenizer(reader.readLine());
        int n = Integer.parseInt(stringTokenizer.nextToken());
        int k = Integer.parseInt(stringTokenizer.nextToken());
        List<List<Edge>> graph = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            graph.add(new ArrayList<>());
        }
        for (int i = 0; i < n - 1; i++) {
            stringTokenizer = new StringTokenizer(reader.readLine());
            int from = Integer.parseInt(stringTokenizer.nextToken()) - 1;
            int to = Integer.parseInt(stringTokenizer.nextToken()) - 1;
            int c = Integer.parseInt(stringTokenizer.nextToken());
            graph.get(from).add(new Edge(to, c));
            graph.get(to).add(new Edge(from, c));
        }
        boolean[] visited = new boolean[n];
        long ans = 0;
        for (int i = 0; i < n; i++) {
            if (visited[i]) continue;
            res = 0;
            dfs(graph, visited, i);
            ans += pow(res, k);
        }

        long possibleans = (pow(n, k) - ans) % mod;
        if (possibleans < 0) {
            possibleans += mod;
        }
        System.out.println(possibleans);

    }
}
