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;
using LL = long long;
constexpr int maxn = 240000;
constexpr LL mod = 1000000007;
int f[maxn];
int getf(int u) { return f[u] < 0 ? u : f[u] = getf(f[u]); }
LL power(LL a, LL r) {
  LL res = 1;
  for (; r; r >>= 1, a = a * a % mod)
    if (r & 1) res = res * a % mod;
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i += 1) f[i] = -1;
  for (int i = 1; i < n; i += 1) {
    int u, v, x;
    cin >> u >> v >> x;
    if (x == 0) {
      u = getf(u);
      v = getf(v);
      if (f[u] < f[v])
        f[u] += f[v], f[v] = u;
      else
        f[v] += f[u], f[u] = v;
    }
  }
  LL ans = power(n, k);
  for (int i = 1; i <= n; i += 1)
    if (f[i] < 0) ans = (ans - power(-f[i], k) + mod) % mod;
  cout << ans;
}
