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;
long long ans = 0;
vector<int> groups;
void mainRecur(vector<vector<pair<int, int>>> &tree, int currentVertex,
               int posToAdd, int parent) {
  groups[posToAdd]++;
  for (int i = 0; i < tree[currentVertex].size(); i++) {
    if (tree[currentVertex][i].first != parent) {
      if (tree[currentVertex][i].second == 1) {
        groups.push_back(0);
        mainRecur(tree, tree[currentVertex][i].first, groups.size() - 1,
                  currentVertex);
      } else {
        mainRecur(tree, tree[currentVertex][i].first, posToAdd, currentVertex);
      }
    }
  }
}
int power(int x, unsigned int y, int p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  int m = 1000000007;
  vector<vector<pair<int, int>>> v;
  int n, k;
  cin >> n >> k;
  for (int i = 0; i <= n; i++) {
    vector<pair<int, int>> temp;
    v.push_back(temp);
  }
  groups.push_back(0);
  for (int i = 0; i < n - 1; i++) {
    int s, d, c;
    cin >> s >> d >> c;
    v[s].push_back(make_pair(d, c));
    v[d].push_back(make_pair(s, c));
  }
  mainRecur(v, 1, 0, -1);
  int ans = 0;
  int forIntital = 1;
  for (int i = 0; i < k; i++) {
    forIntital = ((long long)forIntital * n) % m;
  }
  ans = forIntital;
  for (int i = 0; i < groups.size(); i++) {
    int temp = power(groups[i], k, m);
    ans -= temp;
    if (ans < 0) ans += m;
  }
  cout << ans << endl;
  return 0;
}
