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;
const long long int N = 1000005;
const long long int LG = 22;
vector<long long int> adj[N];
long long int Euler[2 * N + 100];
long long int vis[N];
map<pair<long long int, long long int>, long long int> deep;
void eulerTree(long long int u, long long int &indx) {
  vis[u] = 1;
  Euler[indx++] = u;
  for (auto it : adj[u]) {
    if (!vis[it]) {
      eulerTree(it, indx);
      Euler[indx++] = u;
    }
  }
}
vector<long long int> v;
void printEulerTour(long long int root, long long int n) {
  long long int index = 0;
  eulerTree(root, index);
  for (long long int i = 0; i < (2 * n - 1); i++) v.push_back(Euler[i]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  long long int k;
  cin >> k;
  for (long long int i = 0; i < n - 1; i++) {
    long long int u, vv, col;
    cin >> u >> vv >> col;
    adj[u].push_back(vv);
    adj[vv].push_back(u);
    if (col == 0) deep[make_pair(u, vv)] = deep[make_pair(vv, u)] = 1;
  }
  set<long long int> s;
  printEulerTour(1, n);
  long long int ans = 1;
  for (long long int i = 1; i <= k; i++) ans = (ans * n) % 1000000007;
  vector<long long int> kiki;
  map<long long int, long long int> mapi;
  long long int l = 1;
  set<long long int> yo;
  for (long long int i = 1; i < v.size(); i++) {
    long long int x = v[i], y = v[i - 1];
    if (deep[make_pair(x, y)])
      yo.insert(x), yo.insert(y), mapi[x] = 1, mapi[y] = 1;
    else
      l = 1, kiki.push_back(yo.size()), yo.clear();
  }
  if (yo.size() > 1) kiki.push_back(yo.size());
  for (auto i : kiki) {
    long long int temp = 1;
    for (long long int j = 1; j <= k; j++) temp = (temp * i) % 1000000007;
    ans = (ans - temp + 1000000007) % 1000000007;
  }
  long long int lol = 0;
  for (long long int i = 1; i <= n; i++)
    if (mapi[i] == 0) lol++;
  ans = (ans - lol + 1000000007) % 1000000007;
  cout << ans << endl;
  return 0;
}
