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;
long long md = 1e9 + 7;
const int N = 200005;
int INF = 1e9 + 7;
int n, k;
vector<vector<int>> g1(200005), g0(200005);
bool used[200005];
long long ans;
long long dfs(int v) {
  used[v] = 1;
  queue<int> q;
  q.push(v);
  long long ans1 = 1;
  while (!q.empty()) {
    int t = q.front();
    q.pop();
    for (int i = 0; i < g0[t].size(); i++)
      if (!used[g0[t][i]]) {
        used[g0[t][i]] = 1;
        ans1++;
        q.push(g0[t][i]);
      }
  }
  return ans1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i <= n; i++) used[i] = 0;
  for (int i = 0; i < n - 1; i++) {
    int x, y, c;
    cin >> x >> y >> c;
    if (c) {
      g1[x].push_back(y);
      g1[y].push_back(x);
    } else {
      g0[x].push_back(y);
      g0[y].push_back(x);
    }
  }
  ans = 1;
  for (int i = 0; i < k; i++) {
    ans *= (long long)n;
    ans %= md;
  }
  for (int i = 1; i <= n; i++)
    if (!used[i]) {
      long long n1 = dfs(i);
      if (n1 == 1) {
        ans--;
        if (ans < 0) ans += md;
        ans %= md;
      } else {
        long long ans1 = 1;
        for (int j = 0; j < k; j++) {
          ans1 *= n1;
          ans1 %= md;
        }
        ans -= ans1;
        if (ans < 0) ans += md;
        ans %= md;
      }
    }
  cout << ans;
  return 0;
}
