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>
const int N = 1e6;
const int base = 1e9 + 7;
using namespace std;
int n, k, check[N], cnt[N], res;
vector<int> g[N], co[N];
void dfs(int u, int c) {
  check[u] = true;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i], c = co[u][i];
    if (!check[v]) {
      dfs(v, c);
      if (!c) cnt[u] += cnt[v];
    }
  }
  if (c) {
    int t = 1;
    for (int i = 1; i <= k; i++) t = (t * cnt[u]) % base;
    res += t;
  }
}
void solve() {
  cin >> n >> k;
  for (int i = 1; i < n; i++) {
    int u, v, c;
    cin >> u >> v >> c;
    g[u].push_back(v);
    co[u].push_back(c);
    g[v].push_back(u);
    co[v].push_back(c);
  }
  for (int i = 1; i <= n; i++) cnt[i] = 1;
  dfs(1, 1);
  int ans = 1;
  for (int i = 1; i <= k; i++) ans *= n;
  cout << (ans + base - res) % base;
}
int main() {
  solve();
  return 0;
}
