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;
class Solution {
 public:
  static constexpr long long MOD = 1000000007LL;
  vector<vector<int>> graph;
  array<bool, 100000> used;
  void reset() {
    graph.clear();
    used.fill(false);
  }
  int dfs(int v) {
    used[v] = true;
    int count = 1;
    for (int next : graph[v]) {
      if (used[next]) continue;
      count += dfs(next);
    }
    return count;
  }
  long long binpow(long long a, long long n) {
    long long res = 1;
    while (n) {
      if (n & 1) res = (res * a) % MOD;
      a = (a * a) % MOD;
      n >>= 1;
    }
    return res;
  }
  void solve() {
    reset();
    int n, k;
    cin >> n >> k;
    graph.resize(n);
    for (int i = 0; i < n - 1; ++i) {
      int v, u, b;
      cin >> v >> u >> b;
      v--;
      u--;
      if (!b) {
        graph[v].push_back(u);
        graph[u].push_back(v);
      }
    }
    vector<int> comp_counts;
    for (int v = 0; v < n; ++v) {
      if (!used[v]) {
        comp_counts.push_back(dfs(v));
      }
    }
    long long total = binpow(n, k);
    long long bad = 0;
    for (int count : comp_counts) {
      bad = (bad + binpow(count, k)) % MOD;
    }
    long long result = total - bad;
    if (result < 0) result += MOD;
    cout << result << endl;
  }
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  Solution s;
  int t = 1;
  while (t--) {
    s.solve();
  }
  return 0;
}
