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<int> v[100005];
int vis[100005];
int nodes;
const long long int m = 1e9 + 7;
int dfs(int s) {
  nodes++;
  vis[s] = true;
  for (int i : v[s]) {
    if (!vis[i]) dfs(i);
  }
  return nodes;
}
long long int pow(int x, int n, long long int m) {
  if (n == 0)
    return 1;
  else {
    if (n % 2 == 0)
      return pow((x * x) % m, n / 2, m);
    else
      return x * pow((x * x) % m, (n - 1) / 2, m);
  }
}
int main() {
  int n, k;
  cin >> n >> k;
  int a, b, c;
  for (int i = 1; i <= n; i++) {
    cin >> a >> b >> c;
    if (c == 0) {
      v[a].push_back(b);
      v[b].push_back(a);
    }
  }
  int p;
  long long int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      nodes = 0;
      p = dfs(i);
      ans += pow(p, k, m);
      ans %= m;
    }
  }
  long long int fans = pow(n, k, m);
  fans = (fans - ans + m) % m;
  cout << fans;
  return 0;
}
