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;
vector<vector<pair<long long, long long> > > grafo;
vector<int> tomados;
int n, k;
long long total;
void dfs(int u) {
  tomados[u] = 1;
  for (int i = 0; i < (int)grafo[u].size(); ++i) {
    pair<long long, long long> v = grafo[u][i];
    if (!tomados[v.first]) {
      if (v.second) continue;
      dfs(v.first);
      tomados[u] += tomados[v.first];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  vector<pair<long long, long long> > empty;
  cin >> n >> k;
  grafo.assign(n, empty);
  tomados.assign(n, 0);
  int x, y, z;
  for (int i = 0; i < n - 1; ++i) {
    cin >> x >> y >> z;
    grafo[x - 1].push_back({y - 1, z});
    grafo[y - 1].push_back({x - 1, z});
  }
  total = 1;
  for (int i = 0; i < k; ++i) {
    total *= n;
    total %= 1000000000 + 7;
  }
  for (int i = 0; i < n; ++i) {
    if (tomados[i]) continue;
    dfs(i);
    long long a = 1;
    for (int j = 0; j < k; ++j) {
      a *= (long long)tomados[i];
      a %= 1000000000 + 7;
    }
    total -= a;
    total += 1000000000 + 7;
    total %= 1000000000 + 7;
  }
  cout << total << '\n';
  return 0;
}
