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<vector<long long>> g;
vector<bool> visited;
long long dfs(long long second) {
  visited[second] = 1;
  long long nov = 1;
  for (long long i = 0; i < g[second].size(); i++) {
    if (visited[g[second][i]] == 0) nov += dfs(g[second][i]);
  }
  return nov;
}
int main() {
  long long n, k, u, v, b;
  cin >> n >> k;
  g.resize(n + 1, vector<long long>());
  visited.resize(n + 1, 0);
  for (long long i = 1; i <= n - 1; i++) {
    cin >> u >> v >> b;
    if (b == 0) {
      g[u].push_back(v);
      g[v].push_back(u);
    }
  }
  for (long long i = 1; i <= n; i++) {
    cout << endl << i << " = ";
    for (long long j = 0; j < g[i].size(); j++) cout << g[i][j] << " ";
  }
  vector<long long> ans;
  for (long long i = 1; i <= n; i++) {
    if (visited[i] == 0) ans.push_back(dfs(i));
  }
  cout << endl;
  for (long long i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  long long sum = 0;
  for (long long i = 0; i < ans.size(); i++) {
    long long tk = k, mul = 1;
    while (tk--) mul = (mul * ans[i]) % 1000000007;
    sum = (sum + mul) % 1000000007;
  }
  long long tmul = 1;
  while (k--) tmul = (tmul * n) % 1000000007;
  cout << (tmul - sum) % 1000000007;
}
