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. 

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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 + 7;
vector<pair<int, int>> v(N);
long long modexp(long long x, long long n, long long m) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return modexp((x * x) % m, n / 2, m);
  else
    return (x * modexp((x * x) % m, (n - 1) / 2, m)) % m;
}
int find(int i) {
  if (v[i].first != i) v[i].first = find(v[i].first);
  return v[i].first;
}
void unions(int x, int y) {
  int px = find(x);
  int py = find(y);
  if (px == py) return;
  if (v[px].second > v[py].second) {
    v[py].first = px;
    v[px].second += v[py].second;
  } else {
    v[px].first = py;
    v[py].second += v[px].second;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long i, j;
    for (int i = 1; i <= n; i++) {
      v[i].first = i;
      v[i].second = 1;
    }
    long long x, y, w;
    for (int i = 0; i < n - 1; i++) {
      cin >> x >> y >> w;
      if (w == 0) {
        unions(x, y);
      }
    }
    long long ans = modexp(n, k, 1000000007) - n;
    long long sm = 0;
    for (int i = 1; i < n + 1; i++) {
      if (v[i].first == i) {
        long long sz = v[i].second;
        long long tmp = modexp(sz, k, 1000000007) - sz;
        sm = (sm + tmp) % 1000000007;
      }
    }
    ans = ((ans % 1000000007) - (sm % 1000000007)) % 1000000007;
    cout << ans << '\n';
  }
  return 0;
}
