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 int INF = 1000000000;
const int MAX = 100005;
const int MAX1 = 1005;
const int MAX2 = 105;
const int LEN = 105;
const int BASE = 1000000000;
const double EPS = 1e-7;
const int MOD = 1000000007;
int par[100005];
int pow(int a, int b, int m) {
  int ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
int root(int v) {
  if (par[v] < 0)
    return v;
  else
    return par[v] = root(par[v]);
}
void connect(int u, int v) {
  int n = root(u), m = root(v);
  if (n == m) return;
  if (par[n] < par[m]) {
    par[n] += par[m];
    par[m] = n;
  } else {
    par[m] += par[n];
    par[n] = m;
  }
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ifstream in("");
  ofstream out("");
  clock_t begin = clock();
  for (int i = (0); i < (100005); ++i) par[i] = -1;
  int n, k;
  cin >> n >> k;
  for (int i = (0); i < (n - 1); ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    if (c == 0) {
      connect(a, b);
    }
  }
  long long res = 0;
  set<int> S;
  for (int i = (1); i < (n + 1); ++i) {
    int r = root(i);
    if (S.find(r) != S.end()) continue;
    S.insert(r);
    res += pow(-1 * par[r], k, MOD);
  }
  cout << (pow(n, k, MOD) - res) % MOD << '\n';
  clock_t end = clock();
  double esecs = double(end - begin) / CLOCKS_PER_SEC;
  return 0;
}
