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 long long N = 100100;
const long long Mod = 1e9 + 7;
bool Vis[N];
vector<long long> Adj[N];
void sub(long long& x, long long y) { x = (x - y + 10 * Mod) % Mod; }
long long fp(long long x, long long y) {
  if (!y) return 1;
  long long Res = fp(x, y / 2);
  Res = (Res * Res) % Mod;
  if (y & 1) Res = (x * Res) % Mod;
  return Res;
}
long long DFS(long long i) {
  if (Vis[i]) return 0;
  Vis[i] = true;
  long long Res = 1;
  for (auto j : Adj[i]) Res += DFS(j);
  return Res;
}
int main() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 1; i < n; ++i) {
    long long x, y, w;
    scanf("%I64d%I64d%I64d", &x, &y, &w);
    if (w) continue;
    Adj[x].push_back(y);
    Adj[y].push_back(x);
  }
  long long Ans = fp(n, k);
  for (long long i = 1; i <= n; i++)
    if (!Vis[i]) sub(Ans, fp(DFS(i), k));
  cout << Ans;
}
