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 int res = 0;
long long int checker[100005] = {0};
vector<long long int> v[100005];
long long int power(long long int n, long long int k) {
  long long int result = 1;
  while (k > 0) {
    if (k % 2 == 1) result = (result * n) % 1000000007;
    n = (n * n) % 1000000007;
    k /= 2;
  }
  return result % 1000000007;
}
void dfs(long long int x) {
  if (checker[x] == 1) return;
  res++;
  checker[x] = 1;
  long long int i;
  for (i = 0; i < v[x].size(); i++) {
    dfs(v[x][i]);
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    long long int n, k, i, j;
    cin >> n >> k;
    for (i = 0; i < n - 1; i++) {
      long long int a, b, c;
      cin >> a >> b >> c;
      if (c == 0) {
        v[a].push_back(b);
        v[b].push_back(a);
      }
    }
    long long int ans = power(n, k);
    for (i = 1; i <= n; i++) {
      if (checker[i] == 1) continue;
      res = 0;
      dfs(i);
      long long int sub = power(res, k);
      ans -= sub;
      ans += 1000000007;
      ans = ans % 1000000007;
    }
    cout << ans << endl;
  }
  return 0;
}
