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 nod(long long a, long long b) {
  while (a != 0 && b != 0) (a > b) ? a %= b : b %= a;
  return a + b;
}
long long nok(long long a, long long b) { return (a * b) / nod(a, b); }
const double pi = acos(-1);
const int mod = 1e9 + 7;
const int inf = 2e9;
const int maxlen = 2e5;
struct dot {
  double x, y;
  dot() {}
  dot(double x, double y) : x(x), y(y) {}
};
struct line {
  double a, b, c;
  line() {}
};
double d(dot a, dot b) {
  return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
double dist(line l, dot a) {
  return abs(l.a * a.x + l.b * a.y + l.c) / sqrt(l.a * l.a + l.b * l.b);
}
line getline(dot a, dot b) {
  line l;
  l.a = -a.y + b.y;
  l.b = -b.x + a.x;
  l.c = -a.x * b.y + b.x * a.y;
  return l;
}
double segd(dot a, dot b, dot c) {
  double lens = d(b, c), len1 = d(a, b), len2 = d(a, c);
  if ((lens * lens + len1 * len1 - len2 * len2) *
          (lens * lens + len2 * len2 - len1 * len1) >
      0)
    return dist(getline(b, c), a);
  return min(len1, len2);
}
int p[maxlen], sz[maxlen];
void InitDSU(int n) {
  for (int i = 0; i <= n; ++i) sz[i] = 1, p[i] = i;
}
int find(int u) {
  if (p[u] == u) return u;
  return p[u] = find(p[u]);
}
void Union(int u, int v) {
  u = find(u);
  v = find(v);
  if (u == v) return;
  if (sz[u] > sz[v])
    sz[u] += sz[v], p[v] = u;
  else
    sz[v] += sz[u], p[u] = v;
}
int main() {
  int n, k;
  cin >> n >> k;
  InitDSU(n);
  long long ans = 1;
  for (int i = 0; i < n; ++i) {
    int u, v, b;
    cin >> u >> v >> b;
    if (!b) Union(u, v);
  }
  for (int i = 0; i < k; ++i) ans *= n, ans %= mod;
  for (int i = 1; i <= n; ++i)
    if (p[i] == i) {
      long long r = 1;
      for (int j = 0; j < k; ++j) r *= sz[i], r %= mod;
      ans -= r - sz[i];
      ans %= mod;
    }
  cout << (ans - n) % mod;
}
