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;
const long long MAXN = 2e6 + 10, inf = 1e18, mod = 1e9 + 7;
struct pair_hash {
  template <class T1, class T2>
  std::size_t operator()(const std::pair<T1, T2> &p) const {
    auto h1 = std::hash<T1>{}(p.first);
    auto h2 = std::hash<T2>{}(p.second);
    return h1 ^ h2;
  }
};
long long pw(long long te, long long tee) {
  if (tee == 0) return 1;
  long long res = pw(te, tee / 2) % mod;
  return (tee % 2 ? (((te * res) % mod) * res) % mod : (res * res) % mod);
}
long long v, u, c, n, k, cnt, p[MAXN];
bool mark[MAXN];
vector<long long> e[MAXN];
void Read_input() {
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    cin >> v >> u >> c;
    if (c != 1) e[u].push_back(v), e[v].push_back(u);
  }
}
void dfs(long long v, bool f) {
  cnt += f;
  p[cnt]++;
  mark[v] = true;
  for (auto i : e[v])
    if (!mark[i]) dfs(i, false);
}
void Solve() {
  for (long long i = 1; i <= n; i++)
    if (!mark[i]) dfs(i, true);
  long long t = 0;
  for (long long i = 1; i <= cnt; i++) t = (t + (pw(p[i], k) % mod)) % mod;
  cout << (((pw(n, k) - t) % mod + mod) % mod);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  Read_input(), Solve();
  return 0;
}
