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 n, k;
vector<vector<pair<long long, long long> > > nodes;
vector<long long> bad;
int main() {
  cin >> n >> k;
  nodes.resize(n);
  bad.resize(n);
  vector<long long> ins(n);
  for (int i = 0; i < n - 1; i++) {
    long long u, v, x;
    cin >> u >> v >> x;
    u--;
    v--;
    nodes[u].push_back({v, x});
    nodes[v].push_back({u, x});
  }
  queue<long long> q;
  for (int i = 0; i < n; i++) {
    if (nodes[i].size() == 1) {
      q.push(i);
    }
    ins[i] = nodes[i].size();
  }
  vector<bool> used(n);
  while (q.size() > 0) {
    long long cur = q.front();
    q.pop();
    used[cur] = 1;
    bad[cur]++;
    if (ins[cur] == 0) {
      q.push(cur);
      break;
    }
    for (int i = 0; i < nodes[cur].size(); i++) {
      if (used[nodes[cur][i].first]) continue;
      ins[nodes[cur][i].first]--;
      if (nodes[cur][i].second == 0) bad[nodes[cur][i].first] += bad[cur];
      if (ins[nodes[cur][i].first] <= 1) q.push(nodes[cur][i].first);
    }
  }
  used.resize(0);
  used.resize(n);
  while (q.size() > 0) {
    long long cur = q.front();
    q.pop();
    used[cur] = 1;
    for (int i = 0; i < nodes[cur].size(); i++) {
      if (used[nodes[cur][i].first]) continue;
      if (nodes[cur][i].second == 0)
        bad[nodes[cur][i].first] += bad[cur] - bad[nodes[cur][i].first];
      q.push(nodes[cur][i].first);
    }
  }
  long long allAnsw = 0;
  for (int i = 0; i < n; i++) {
    long long answ = 1;
    answ %= (long long)(1e9 + 7);
    long long wrong = 1;
    for (int j = 1; j < k; j++) {
      answ *= n;
      answ %= (long long)(1e9 + 7);
      wrong *= bad[i];
      wrong %= (long long)(1e9 + 7);
    }
    answ = (answ - wrong + (long long)(1e9 + 7)) % (long long)(1e9 + 7);
    allAnsw += answ;
    allAnsw %= (long long)(1e9 + 7);
  }
  cout << allAnsw << endl;
  return 0;
}
