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;
template <class T>
using max_pq = priority_queue<T>;
template <class T>
using min_pq = priority_queue<T, vector<T>, greater<T>>;
int oo = 0x3f3f3f3f;
long long n, k;
vector<vector<pair<int, int>>> graph;
vector<int> redGroups;
set<int> visited;
int dfs(int curr) {
  int size = 1;
  visited.insert(curr);
  for (int i = 0; i < graph[curr].size(); i++) {
    if (graph[curr][i].second == 0 &&
        visited.find(graph[curr][i].first) == visited.end()) {
      size += dfs(graph[curr][i].first);
    }
  }
  return size;
}
class CEdgyTrees {
 public:
  void solve(istream& cin, ostream& cout) {
    graph.clear();
    redGroups.clear();
    cin >> n >> k;
    for (int i = 0; i < n; i++) {
      vector<pair<int, int>> tmp;
      graph.push_back(tmp);
    }
    for (int i = 0; i < n - 1; i++) {
      int u, v, x;
      cin >> u >> v >> x;
      u--;
      v--;
      graph[u].push_back(make_pair(v, x));
      graph[v].push_back(make_pair(u, x));
    }
    for (int i = 0; i < n; i++) {
      if (visited.find(i) == visited.end()) {
        redGroups.push_back(dfs(i));
      }
    }
    long long ans = 1;
    for (int i = 0; i < k; i++) {
      ans = ((ans % 1000000007ll) * n) % 1000000007ll;
    }
    for (int i = 0; i < redGroups.size(); i++) {
      if (redGroups[i] == 1) {
        ans--;
      } else {
        int x = 1;
        for (int j = 0; j < k; j++) {
          x = ((x % 1000000007ll) * (redGroups[i] % 1000000007ll)) %
              1000000007ll;
        }
        ans -= x;
        if (ans < 0) ans += 1000000007ll;
      }
    }
    cout << ans;
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  CEdgyTrees solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}
