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;
void printprecise(double l, long long int precision) {
  std::cout << std::fixed;
  std::cout << std::setprecision(precision);
  std::cout << l;
}
long long int power_wm(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int power(long long int x, long long int y) {
  long long int res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long int root(long long int x, vector<long long int> &roots) {
  if (roots[x] != x) roots[x] = root(roots[x], roots);
  return roots[x];
}
void unoin(long long int p, long long int q, vector<long long int> &roots,
           vector<long long int> &rank) {
  long long int rp = root(p, roots);
  long long int rq = root(q, roots);
  if (rank[rq] < rank[rp]) {
    roots[rq] = rp;
  } else if (rank[rp] > rank[rq]) {
    roots[rp] = rq;
  } else {
    roots[rp] = rq;
    rank[rq]++;
  }
}
long long int dig_sum(long long int n) {
  long long int s = 0;
  while (n) {
    s += n % 10;
    n /= 10;
  }
  return s;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, k;
  cin >> n >> k;
  vector<long long int> roots(n), rank(n);
  for (long long int i = 0; i < n; i++) {
    roots[i] = i;
  }
  long long int a, b, x;
  for (long long int i = 0; i < n - 1; i++) {
    cin >> a >> b >> x;
    if (x) continue;
    a--, b--;
    unoin(a, b, roots, rank);
  }
  vector<long long int> ccs(n);
  for (long long int i = 0; i < n; i++) {
    ccs[root(i, roots)]++;
  }
  long long int badSequences = 0;
  for (long long int i = 0; i < n; i++) {
    badSequences =
        (badSequences + power_wm(ccs[i], k, 1000000007)) % 1000000007;
  }
  cout << (power_wm(n, k, 1000000007) - badSequences + 1000000007) % 1000000007;
  return 0;
}
