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 maxn = 1e5 + 1;
set<long long> a;
const long long mod = 1e9 + 7;
long long exp(long long base, long long expz) {
  long long res = 1;
  while (expz > 0) {
    if (expz % 2 == 1) res = (res * base) % mod;
    base = (base * base) % mod;
    expz /= 2;
  }
  return res % mod;
}
long long inv(long long a) { return exp(a, mod - 2); }
class graph {
 public:
  vector<vector<pair<long long, long long>>> adj;
  vector<long long> parent, level, visited;
  long long size;
  graph(long long n) {
    size = n;
    parent = vector<long long>(n, 0);
    level = vector<long long>(n, 0);
    visited = vector<long long>(n, 0);
    adj = vector<vector<pair<long long, long long>>>(n);
  }
  void addEdge(long long a, long long b, long long v) {
    --a;
    --b;
    adj[a].push_back(make_pair(b, v));
    adj[b].push_back(make_pair(a, v));
  }
  void dfs(long long root, long long parent, long long color = -1) {
    visited[root] = 1;
    if (color == 0) {
      a.insert(root);
      a.insert(parent);
    }
    if (color == 1) {
      return;
    }
    for (auto i : adj[root]) {
      if (!visited[i.first] && i.second == 0) {
        dfs(i.first, root, i.second);
      }
    }
  }
};
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long n, k;
  cin >> n >> k;
  graph t(n);
  for (long long i = 0; i < n - 1; i++) {
    long long a, b, c;
    cin >> a >> b >> c;
    t.addEdge(a, b, c);
  }
  long long res = exp(n, k) - n;
  for (long long i = 0; i < n; i++) {
    if (!t.visited[i]) {
      t.dfs(i, i, -1);
      long long top = a.size();
      res -= exp(top, k) - top;
      a.clear();
      res %= mod;
    }
  }
  if (res < 0) res += mod;
  cout << res << "\n";
  return 0;
}
