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 power(long long a, long long b, long long m) {
  long long ans = 1;
  a %= m;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return ans;
}
vector<long long> adj[100009];
bool vis[100009];
long long sz = 0;
void dfs(int i) {
  vis[i] = true;
  sz++;
  for (auto v : adj[i]) {
    if (vis[v] == false) dfs(v);
  }
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin.tie(0);
  memset(vis, false, sizeof(vis));
  unsigned long long n, k, u, v, c;
  cin >> n >> k;
  for (long long i = 1; i < n; ++i) {
    cin >> u >> v >> c;
    if (c == 0) {
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
  }
  long long ans = (power(n, k, 1000000007) - n) % 1000000007;
  for (long long i = 1; i <= n; ++i) {
    if (vis[i] == false) {
      sz = 0;
      dfs(i);
      ans += 1000000007;
      ans -= power(sz, k, 1000000007);
      ans += sz;
      ans = ans % 1000000007;
    }
  }
  ans = ans % 1000000007;
  while (ans < 0) {
    ans += 1000000007;
  }
  cout << ans;
  return 0;
}
