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 inf = 1e9 + 7, cnt;
map<long long int, vector<long long int>> red, black;
vector<bool> visb, visr;
set<long long int> tmp;
void dfs(long long int r) {
  visr[r] = true;
  tmp.insert(r);
  for (long long int i = 0; i < red[r].size(); ++i) {
    if (visr[red[r][i]] == false and red.find(red[r][i]) != red.end())
      dfs(red[r][i]);
  }
}
void dfsb(long long int r) {
  visb[r] = true;
  tmp.insert(r);
  for (long long int i = 0; i < black[r].size(); ++i) {
    if (visb[black[r][i]] == false and black.find(black[r][i]) != black.end())
      dfsb(black[r][i]);
  }
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  x = x % inf;
  while (y > 0) {
    if (y & 1) res = (res * x) % inf;
    y = y >> 1;
    x = (x * x) % inf;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, k, ans;
  cin >> n >> k;
  ans = power(n, k);
  visb.resize(n + 1, false);
  visr.resize(n + 1, false);
  for (long long int i = 0; i < n - 1; ++i) {
    long long int x, y, z;
    cin >> x >> y >> z;
    if (z == 0)
      red[x].push_back(y), red[y].push_back(x);
    else
      black[x].push_back(y), black[y].push_back(x);
  }
  for (long long int i = (1); i <= (n); ++i) {
    if (red.find(i) != red.end() and visr[i] == 0) {
      tmp.clear();
      dfs(i);
      cnt = tmp.size();
      ans -= power(cnt, k);
      if (ans < 0) ans += inf;
    }
    if (black.find(i) != black.end() and visb[i] == 0) {
      tmp.clear();
      dfsb(i);
      cnt = tmp.size();
      ans -= cnt;
      if (ans < 0) ans += inf;
    }
    if (red.find(i) != red.end() and black.find(i) != black.end()) ans++;
    ans %= inf;
  }
  cout << ans;
  return 0;
}
