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 n, k;
const long long mod = 1e9 + 7;
struct DSU {
  vector<long long> parent;
  vector<long long> rank;
  vector<long long> count;
  DSU(long long n) : parent(n), rank(n), count(n) {
    for (long long i = 0; i < n; ++i) {
      parent[i] = i;
      rank[i] = 0;
      count[i] = 1;
    }
  }
  long long find(long long x) {
    if (parent[x] == x) return x;
    return parent[x] = find(parent[x]);
  }
  void merge(long long x, long long y) {
    x = find(x);
    y = find(y);
    if (rank[x] > rank[y]) swap(x, y);
    parent[x] = y;
    if (rank[x] == rank[y]) rank[y]++;
    count[y] += count[x];
  }
};
long long sqr(long long x) { return (x * x) % mod; }
long long power(long long x, long long k) {
  if (k == 1) return x;
  if (k == 0) return 0;
  if (k % 2 == 0)
    return sqr(power(x, k / 2));
  else
    return (sqr(power(x, k / 2)) * x) % mod;
}
int main() {
  cin >> n >> k;
  long long u, v, x;
  DSU dsu(n);
  for (long long i = 0; i < n - 1; ++i) {
    cin >> u >> v >> x;
    if (x == 0) dsu.merge(u - 1, v - 1);
  }
  long long bad = 0;
  vector<bool> used(n, false);
  for (long long i = 0; i < n; ++i) {
    long long comp = dsu.find(i);
    if (!used[comp]) {
      bad += power(dsu.count[comp], k);
      bad %= mod;
    }
    used[comp] = true;
  }
  long long good = (power(n, k) - bad) % mod;
  cout << good << endl;
  return 0;
}
