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.*;
import java.io.*;
import java.math.*;

public class C {
  public static void main(String[] args) {
    Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
    int n = sc.nextInt();
    int k = sc.nextInt();
    List<List<Integer>> g = new ArrayList<>();
    for (int i = 0; i < n; i++)
      g.add(new ArrayList<>());
    for (int i = 0; i < n-1; i++) {
      int u = sc.nextInt()-1;
      int v = sc.nextInt()-1;
      int x = sc.nextInt();
      if (x==0) {
        g.get(u).add(v);
        g.get(v).add(u);
      }
    }
    List<Integer> sizes = new ArrayList<>();
    ArrayDeque<Integer> q = new ArrayDeque<>();
    boolean[] vis = new boolean[n];
    for (int i = 0; i < n; i++) {
      if (vis[i])
        continue;
      int nn = 0;
      q.add(i);
      vis[i] = true;
      while (!q.isEmpty()) {
        int u = q.removeFirst();
        nn++;
        for (int v : g.get(u)) {
          if (!vis[v]) {
            vis[v] = true;
            q.add(v);
          }
        }
      }
      sizes.add(nn);
    }
    int MOD = 1000000007;
    int[] mods = new int[n+1];
    int ans = 0;
    for (int s : sizes) {
      if (mods[s] > 0) {
        ans += mods[s];
        ans %= MOD;
        continue;
      }
      long ta = 1;
      for (int i = 0; i < k; ++i) {
        ta *= s;
        ta %= MOD;
      }
      mods[s] = (int)ta;
      ans += mods[s];
      ans %= MOD;
    }
    ans = (MOD-ans)%MOD;
    long ta = 1;
    for (int i = 0; i < k; ++i) {
      ta *= n;
      ta %= MOD;
    }
    System.out.println((ta+ans)%MOD);
  }
}
