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.
#include <bits/stdc++.h>
using namespace std;
class DisjSet {
 public:
  int *parent, *count, n;
  DisjSet(int n) {
    parent = new int[n]{0};
    count = new int[n]{0};
    memset(parent, -1, n * sizeof(int));
    this->n = n;
  }
  int _find(int x) {
    if (parent[x] >= 0) {
      parent[x] = _find(parent[x]);
    } else
      return x;
    return parent[x];
  }
  void _union(int x, int y) {
    int parent_x = _find(x);
    int parent_y = _find(y);
    if (parent_x == parent_y) return;
    if (parent_x < parent_y) {
      parent[parent_y] = parent_x;
      count[parent_x] += count[parent_y];
    } else {
      parent[parent_x] = parent_y;
      count[parent_y] += count[parent_x];
    }
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k, u, v, c;
  cin >> n >> k;
  DisjSet ds(n);
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v >> c;
    if (c == 0) {
      if (ds.count[u - 1] == 0) ds.count[u - 1] = 1;
      if (ds.count[v - 1] == 0) ds.count[v - 1] = 1;
      ds._union(u - 1, v - 1);
    }
  }
  unordered_map<int, int> mappy;
  vector<int> count;
  for (int i = 0; i < n; i++) {
    int parent = ds._find(i);
    if (ds.count[parent] != 0 && mappy[parent] == 0) {
      mappy[parent] = 1;
      count.push_back(ds.count[parent]);
    }
  }
  long long int diff = 0, result = 1;
  for (int i = 1; i <= k; i++) {
    if (result > 1e13)
      result = (result * n) % 1000000007;
    else
      result = result * n;
  }
  result -= n;
  for (int i = 0; i < count.size(); i++) {
    long long int temp = 1;
    for (int j = 1; j <= k; j++) temp = (temp * count[i]) % 1000000007;
    temp -= count[i];
    diff = (diff + temp) % 1000000007;
  }
  cout << result - diff;
  return 0;
}
