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<int>> graph;
vector<bool> visited;
const int mod = 1000000007;
long long myPow(long long x, int p) {
  long long result = 1;
  while (true) {
    if (p & 1) {
      result = result * x % mod;
    }
    if (!p) break;
    p = p >> 1;
    x = x * x % mod;
  }
  return result;
}
vector<int> f(int n) {
  vector<int> result;
  for (int i = 0; i < n; i++) {
    if (!visited[i]) {
      int cnt = 0;
      visited[i] = true;
      queue<int> q;
      q.push(i);
      while (!q.empty()) {
        int t = q.front();
        q.pop();
        cnt++;
        for (int j = 0; j != graph[t].size(); j++) {
          if (!visited[graph[t][j]]) {
            visited[graph[t][j]] = true;
            q.push(graph[t][j]);
          }
        }
      }
      result.push_back(cnt);
    }
  }
  return result;
}
int main() {
  int n, k;
  cin >> n >> k;
  graph.resize(n);
  visited.resize(n);
  for (int i = 0; i < n; i++) {
    visited[i] = false;
  }
  for (int i = 0; i < n - 1; i++) {
    int u, v, x;
    cin >> u >> v >> x;
    if (x == 0) {
      graph[u - 1].push_back(v - 1);
      graph[v - 1].push_back(u - 1);
    }
  }
  vector<int> v = f(n);
  long long tmp = 0;
  for (int i = 0; i != v.size(); i++) {
    tmp += myPow(v[i], k);
  }
  long long result = myPow(n, k) - tmp;
  cout << result % mod << endl;
}
