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>
const int N = 1e5 + 2, p = 1e9 + 7;
int siz[N], f[N];
int n, m, x, y, z, i, c;
inline void read(int &x) {
  c = getchar();
  while ((c < 48) || (c > 57)) c = getchar();
  x = c ^ 48;
  c = getchar();
  while ((c >= 48) && (c <= 57)) {
    x = x * 10 + (c ^ 48);
    c = getchar();
  }
}
int getf(register int x) {
  if (f[x] == x) return f[x];
  f[x] = getf(f[x]);
  siz[f[x]] += siz[x];
  siz[x] = 0;
  return f[x];
}
inline int ksm(register int x, register int y) {
  register int r = 1;
  while (y) {
    if (y & 1) r = (long long)r * x % p;
    x = (long long)x * x % p;
    y >>= 1;
  }
  return r;
}
int main() {
  read(n);
  read(m);
  for (i = 1; i <= n; i++) siz[f[i] = i] = 1;
  for (i = 1; i < n; i++) {
    read(x);
    read(y);
    read(z);
    if (z == 0) {
      siz[getf(y)] += siz[getf(x)];
      siz[f[x]] = 0;
      f[f[x]] = f[y];
    }
  }
  c = ksm(n, m);
  for (i = 1; i <= n; i++)
    if (siz[i]) c = (c - ksm(siz[i], m) + p) % p;
  printf("%d", c);
}
