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 mod = 1000000007;
int pow(long long x, long long n) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % mod;
    x = (x * x) % mod;
    n = n / 2;
  }
  return result;
}
int main() {
  long long n, k;
  cin >> n >> k;
  long long parent[n + 1];
  long long count[n + 1];
  memset(parent, -1, sizeof(parent));
  memset(count, 0, sizeof(count));
  for (long long i = 0; i < n - 1; i++) {
    long long u, v, x;
    cin >> u >> v >> x;
    if (x == 0) {
      if (parent[u] == -1 && parent[v] == -1) {
        parent[v] = u;
        parent[u] = u;
        count[u] += 2;
      } else if (parent[v] != -1) {
        parent[u] = parent[v];
        count[parent[v]]++;
      } else if (parent[u] != -1) {
        parent[v] = parent[u];
        count[parent[u]]++;
      }
    }
  }
  set<long long> s;
  long long si = 0;
  for (long long i = 1; i <= n; i++) {
    if (parent[i] == i) {
      s.insert(i);
    } else if (parent[i] == -1) {
      si++;
    }
  }
  long long tobesub = 0;
  set<long long>::iterator itr;
  for (itr = s.begin(); itr != s.end(); itr++) {
    tobesub += pow(count[*itr], k);
  }
  long long ans = ((pow(n, k) - tobesub - si) + mod) % mod;
  cout << ans << endl;
}
