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;
int dsu_root(int a, int *parent) {
  return (a == parent[a]) ? a : parent[a] = dsu_root(parent[a], parent);
}
void dsu_join(int a, int b, int *parent) {
  parent[dsu_root(b, parent)] = dsu_root(a, parent);
}
constexpr int MOD = 1e9 + 7;
unsigned pow_mod(unsigned b, unsigned p) {
  uint64_t r = 1;
  uint64_t b2 = b;
  while (p) {
    if (p & 1) {
      r *= b2;
      r %= MOD;
    }
    b2 *= b2;
    b2 %= MOD;
    p >>= 1;
  }
  return unsigned(r);
}
int main(int argc, char **argv) {
  int n, k;
  scanf("%d%d", &n, &k);
  int parent[n + 1];
  memset(parent, 0, sizeof(parent));
  for (int i = 0; i < n - 1; i++) {
    int u, v, x;
    scanf("%d%d%d", &u, &v, &x);
    if (!x) {
      if (!parent[u]) parent[u] = u;
      if (!parent[v]) parent[v] = v;
      dsu_join(u, v, parent);
    }
  }
  int sz[n + 1];
  memset(sz, 0, sizeof(sz));
  for (int i = 1; i <= n; i++) sz[dsu_root(i, parent)]++;
  unsigned ans = pow_mod(n, k);
  ans += MOD - n;
  ans %= MOD;
  for (int i = 1; i <= n; i++) {
    if (sz[i] > 0) {
      unsigned minus = pow_mod(sz[i], k);
      minus += MOD - sz[i];
      minus %= MOD;
      ans += MOD - minus;
      ans %= MOD;
    }
  }
  printf("%u\n", ans);
  return 0;
}
