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;
using namespace std::chrono;
const int N = 1e6, M = N, inf = (int)1e9;
const long long mod = 1e9 + 7;
int fastAbs(int n) { return (n ^ (n >> 31)) - (n >> 31); }
long long multiply(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long add(long long a, long long b) {
  return ((a % mod) + (b % mod)) % mod;
}
long long sub(long long a, long long b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
int ans;
void read(vector<int>& a) {
  for (int i = 0; i < a.size(); i++) cin >> a[i];
}
map<pair<int, int>, int> mp;
const int dx[4] = {1, 0, 0, -1};
const int dy[4] = {0, -1, 1, 0};
void solve() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  map<int, int> mp1, mp0, common;
  for (int i = 0; i < n - 1; i++) {
    int x, y, color;
    cin >> x >> y >> color;
    if (color == 1) {
      mp1[x]++;
      mp1[y]++;
    } else {
      if (mp1.find(x) != mp1.end()) {
        common[x]++;
      }
      if (mp1.find(y) != mp1.end()) {
        common[y]++;
      }
      mp0[x]++;
      mp0[y]++;
    }
  }
  long long A = mp0.size();
  long long B = mp1.size();
  long long ans = 1;
  for (int i = 0; i < k; i++) {
    ans = multiply(ans, n);
  }
  long long sub1 = 1;
  for (int i = 0; i < k; i++) {
    sub1 = multiply(sub1, A);
  }
  ans = sub(ans, sub1);
  ans = sub(ans, B);
  ans = add(ans, (long long)common.size());
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int q = 1;
  while (q--) solve();
}
