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>
#pragma GCC optimize("O3")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long rand_seed() {
  long long a = rng();
  return a;
}
int n, k;
long long nr[102][100002], gg[102][100002];
vector<pair<int, int> > v[100002];
vector<int> cc[100002];
int ccc;
bool viz[100002];
void dfs(int nod) {
  cc[ccc].push_back(nod);
  viz[nod] = 1;
  for (int i = 0; i < v[nod].size(); ++i) {
    int vecin = v[nod][i].first;
    int cl = v[nod][i].second;
    if (cl == 0 && !viz[vecin]) dfs(vecin);
  }
}
long long tsum[100002];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 1; i < n; ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    v[a].push_back({b, c});
    ;
    v[b].push_back({a, c});
  }
  for (int i = 1; i <= n; ++i) nr[1][i] = 1;
  for (int i = 1; i <= n; ++i) {
    if (!viz[i]) {
      ++ccc;
      dfs(i);
    }
  }
  for (int i = 1; i < k; ++i) {
    long long totsum = 0;
    long long tt = 0;
    for (int j = 1; j <= ccc; ++j) {
      long long sum = 0;
      for (int p = 0; p < cc[j].size(); ++p)
        sum = (sum + nr[i][cc[j][p]]) % 1000000007,
        tt = (tt + gg[i][cc[j][p]]) % 1000000007;
      for (int p = 0; p < cc[j].size(); ++p) nr[i + 1][cc[j][p]] = sum;
      totsum = (totsum + sum) % 1000000007;
      tsum[j] = sum;
    }
    for (int j = 1; j <= ccc; ++j) {
      long long sum = totsum + 1000000007 - tsum[j];
      sum %= 1000000007;
      for (int p = 0; p < cc[j].size(); ++p)
        gg[i + 1][cc[j][p]] = (sum + tt) % 1000000007;
    }
  }
  long long ans = 0;
  for (int j = 1; j <= n; ++j) ans = (ans + gg[k][j]) % 1000000007;
  cout << ans;
  return 0;
}
