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 int M = 1000000007;
bool check[100010] = {false};
vector<long long int> cou;
long long int c = 0;
void addEdge(vector<long long int> adj[], long long int u, long long int v) {
  adj[u].push_back(v);
  adj[v].push_back(u);
}
void DFS(vector<long long int> adj[], long long int s) {
  check[s] = true;
  for (long long int i = 0; i < adj[s].size(); i++) {
    if (check[adj[s][i]] == false) DFS(adj, adj[s][i]);
  }
  c++;
}
long long int power(long long int a, long long int n) {
  long long int result = 1, x = a;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result % M;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, k;
  cin >> n >> k;
  long long int sum = power(n, k);
  vector<long long int> adj[n + 1];
  for (long long int i = 0; i < n - 1; i++) {
    long long int u, v, b;
    cin >> u >> v >> b;
    if (b == 0) addEdge(adj, u, v);
  }
  for (long long int i = 1; i <= n; i++) {
    if (check[i] == false) {
      c = 0;
      DFS(adj, i);
      cou.push_back(c);
    }
  }
  for (long long int i = 0; i < cou.size(); i++) {
    if (0 < cou[i] && cou[i] <= n) sum = (sum % M - power(cou[i], k) % M) % M;
  }
  cout << sum % M;
  return 0;
}
