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>
#pragma GCC optimize("O3")
using namespace std;
vector<int> rk, f;
vector<long long> sum;
set<int> s;
int find(int x) { return f[x] == x ? x : f[x] = find(f[x]); }
void merge(int x, int y) {
  if (rk[x] >= rk[y]) {
    f[y] = x;
    rk[x] += rk[y];
  } else {
    f[x] = y;
    rk[y] += rk[x];
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, k, a;
  int b, c, cnt;
  f.reserve(200000);
  rk.reserve(200000);
  sum.reserve(100000);
  cin >> n >> k;
  cnt = 0;
  rk.resize(n, 1);
  f.resize(n);
  for (int i = 0; i < n; i++) {
    f[i] = i;
  }
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b >> c;
    if (!c) {
      cnt++;
      a = find(a);
      b = find(b);
      if (a != b) merge(a, b);
    }
  }
  if (cnt == n - 1) {
    cout << 0 << '\n';
    return 0;
  }
  for (int i = 0; i < n; i++) {
    a = find(i);
    if (rk[a] > 1 && !s.count(a)) {
      s.insert(a);
      sum.push_back(rk[a]);
    }
  }
  long long ans = n % 1000000007;
  for (int i = 1; i < k; i++) {
    ans *= n;
    ans %= 1000000007;
  }
  ans -= n;
  for (int i = 0; i < sum.size(); i++) {
    long long tmp = sum[i];
    for (int j = 1; j < k; j++) {
      tmp *= sum[i];
      tmp %= 1000000007;
    }
    tmp -= sum[i];
    ans -= tmp;
    if (ans < 0) ans += 1000000007;
  }
  cout << ans % 1000000007 << '\n';
  return 0;
}
