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 n, k, i, j, x, rez, nr, y, ok, md = 1000000007;
long long z[100005];
struct per {
  long long x, t;
} s;
vector<per> ad[100005];
deque<per> q;
int main() {
  cin >> n >> k;
  for (i = 1; i < n; i++) {
    cin >> x >> y >> j;
    ad[x].push_back({y, j});
    ad[y].push_back({x, j});
    if (j == 1) ok = 1;
  }
  rez = 1;
  for (i = 1; i <= k; i++) rez = rez * n % md;
  q.push_back({1, 0});
  z[s.x] = 1;
  while (!q.empty()) {
    s = q.front();
    q.pop_front();
    z[s.x] = 1;
    if (s.t == 0)
      nr++;
    else {
      long long sc = 1;
      for (i = 1; i <= k; i++) sc = sc * nr % md;
      rez -= sc;
      if (rez < 0) rez += md;
      nr = 1;
    }
    for (auto i : ad[s.x]) {
      if (z[i.x] == 0) {
        z[i.x] = 1;
        if (i.t == 0)
          q.push_front(i);
        else
          q.push_back(i);
      }
    }
  }
  if (nr) {
    long long sc = 1;
    for (i = 1; i <= k; i++) sc = sc * nr % md;
    rez -= sc;
    if (rez < 0) rez += md;
  }
  if (ok == 0) rez = 0;
  cout << rez << "\n";
}
