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 N, K, newp = 0;
int first[200000 + 5], vis[200000 + 5] = {0};
int size[200000 + 5] = {0}, scc = 0;
long long ans = 1, num = 1;
struct wc {
  int to;
  int ne;
} v[200000 * 2 + 5];
void add(int a, int b) {
  v[++newp] = (wc){b, first[a]};
  first[a] = newp;
}
void dfs(int root) {
  vis[root] = 1, size[scc] += 1;
  for (int i = first[root]; i; i = v[i].ne) {
    int j = v[i].to;
    if (vis[j]) continue;
    dfs(j);
  }
}
int main() {
  scanf("%d%d", &N, &K);
  for (int i = 1, x, y, c; i < N; i++) {
    scanf("%d%d%d", &x, &y, &c);
    if (c == 0) add(x, y);
  }
  for (int i = 1; i <= N; i++)
    if (!vis[i]) scc++, dfs(i);
  for (int i = 1; i <= K; i++) ans = ans * N, ans %= 1000000007;
  ans -= N;
  for (int i = 1; i <= scc; i++) {
    num = 1;
    for (int j = 1; j <= K; j++) {
      num *= size[i];
      num %= 1000000007;
    }
    num -= size[i];
    ans = (ans - num + 1000000007) % 1000000007;
  }
  printf("%lld", ans);
  return 0;
}
