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 i64 = long long;
template <class T>
T pow(T x, T n, T mod) {
  T ret = 1;
  while (n > 0) {
    if ((n & 1) == 1) ret = (ret * x) % mod;
    x = (x * x) % mod;
    n >>= 1;
  }
  return ret;
}
template <class T>
T inverse(T x, T p) {
  return pow(x, p - 2, p);
}
template <class T = int>
class UF {
  T *par;

 public:
  UF(T n) : par(new T[n]) {
    for (T i = 0; i < n; i++) par[i] = i;
  }
  ~UF() { delete[] par; }
  T root(T x) {
    if (par[x] == x) return x;
    return par[x] = root(par[x]);
  }
  bool same(T x, T y) { return root(x) == root(y); }
  void unite(T x, T y) {
    x = root(x);
    y = root(y);
    if (x != y) par[x] = y;
  }
};
constexpr i64 mod = 1000000007;
std::vector<std::deque<std::pair<int, int>>> g;
int main() {
  i64 n, k;
  scanf("%lld%lld", &n, &k);
  g.resize(n);
  UF<> uft(n);
  std::vector<int> s(n, 1);
  for (int i = 1; i < n; i++) {
    int u, v, x;
    scanf("%d%d%d", &u, &v, &x);
    u--;
    v--;
    g[u].emplace_back(v, x);
    g[v].emplace_back(u, x);
    if (!x) {
      uft.unite(u, v);
    }
  }
  std::map<int, std::deque<int>> m;
  for (int i = 0; i < n; i++) {
    m[uft.root(i)].push_back(i);
  }
  i64 ret = 0;
  for (auto &el : m) {
    i64 size = el.second.size();
    ret = (ret + pow(size, k, mod)) % mod;
  }
  ret = (pow(n, k, mod) + mod - ret) % mod;
  std::cout << ret << std::endl;
  return 0;
}
