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;
int64_t MOD = 1000000007;
int32_t DFS(int32_t v, vector<vector<int32_t>>& G, vector<bool>& F) {
  F[v] = true;
  int32_t answer = 1;
  for (auto w : G[v]) {
    if (!F[w]) {
      answer += DFS(w, G, F);
    }
  }
  return answer;
}
int64_t POW(int64_t x, int32_t a) {
  if (a == 0) {
    return 0;
  }
  if (a == 1) {
    return x;
  }
  if (a % 2 == 1) {
    return (x * POW(x, a - 1)) % MOD;
  }
  int64_t tmp = POW(x, a / 2);
  return (tmp * tmp) % MOD;
}
int main() {
  int32_t n, k;
  cin >> n >> k;
  vector<vector<int32_t>> G(n);
  for (int32_t _ = 0; _ < n - 1; ++_) {
    int32_t u, v, x;
    cin >> u >> v >> x;
    --u;
    --v;
    if (x == 0) {
      G[u].push_back(v);
      G[v].push_back(u);
    }
  }
  vector<bool> F(n, false);
  int64_t answer = 0;
  for (int32_t i = 0; i < n; ++i) {
    if (!F[i]) {
      int64_t current = static_cast<int64_t>(DFS(i, G, F));
      answer += POW(current, k);
      answer %= MOD;
    }
  }
  answer = (MOD + POW(n, k) - answer) % MOD;
  cout << answer << endl;
  return 0;
}
