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;
const long long mod = 1e9 + 7;
const long long N5 = 1e5 + 9;
const long long N6 = 1e6 + 9;
long long pow(long long a, unsigned long long b, long long m) {
  long long ans = 1;
  a = a % m;
  while (b > 0) {
    if (b & 1) {
      ans = (ans * a) % m;
    }
    a = (a * a) % m;
    b = b >> 1;
  }
  return ans;
}
long long modInverse(long long a, long long m) { return pow(a, m - 2, m) % m; }
void init() {}
long long ans = 0;
void DFS(vector<long long> graph[], long long u, long long visited[]) {
  ans++;
  visited[u] = 1;
  for (long long i = 0; i < graph[u].size(); i++) {
    if (visited[graph[u][i]] == 0) DFS(graph, graph[u][i], visited);
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, k;
  cin >> n >> k;
  vector<long long> graph[n];
  for (long long i = 0; i < n - 1; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    if (w == 0) {
      graph[u - 1].push_back(v - 1);
      graph[v - 1].push_back(u - 1);
    }
  }
  long long visited[n];
  memset(visited, 0, sizeof(visited));
  long long total = pow(n, k, mod);
  for (long long i = 0; i < n; i++) {
    if (visited[i] == 0) {
      ans = 0;
      DFS(graph, i, visited);
      total = (total - pow(ans, k, mod) + mod) % mod;
    }
  }
  cout << total << endl;
  return 0;
}
