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;
const int N = 1e5 + 1;
const long long mod = 1e9 + 7;
const int inf = 0x7fffffff;
int shuyu[N], num[N], fa[N];
bool flag[N];
int n, k, t = 0;
int find(int x) {
  if (fa[x] == x) return x;
  return fa[x] = find(fa[x]);
}
int poww(int a, long long n, int p) {
  int x = 1;
  while (n) {
    if (n & 1) x = 1ll * x * a % p;
    a = 1ll * a * a % p;
    n >>= 1;
  }
  return x;
}
int main() {
  scanf("%d%d", &n, &k);
  int g, j, m;
  for (int i = 0; i < N; i++) fa[i] = i;
  for (int i = 1; i < n; i++) {
    scanf("%d%d%d", &g, &j, &m);
    if (!m) {
      fa[find(g)] = find(j);
      flag[g] = flag[j] = true;
    }
  }
  for (int i = 1; i <= n; i++)
    if (flag[i]) {
      shuyu[i] = find(i);
      num[shuyu[i]]++;
    }
  long long ans = poww(n, k, mod);
  for (int i = 1; i <= n; i++) {
    if (num[i]) ans = 1ll * (ans - poww(num[i], k, mod) + mod) % mod;
    if (!shuyu[i]) ans = 1ll * (ans - 1 + mod) % mod;
  }
  printf("%I64d\n", ans % mod);
  return 0;
}
