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;
vector<int> v[100005];
long long int vis[100005] = {0}, k, a[100005], z, ans, x = 0;
map<pair<int, int>, int> m;
int bigmod(long long int a, long long int b, long long int m) {
  if (b == 0) {
    return 1 % m;
  }
  long long int x = bigmod(a, b / 2, m);
  x = (x * x) % m;
  if (b % 2 == 1) x = (x * a) % m;
  return x;
}
void dfs(long long int x) {
  long long int i;
  vis[x] = 1;
  for (i = 0; i < v[x].size(); i++) {
    long long int d = v[x][i];
    if (vis[d] == 0) {
      dfs(d);
      if (m[make_pair(x, d)] == 0) {
        a[x] += a[d];
      }
      if (m[make_pair(x, d)] == 1) {
        ans -= (bigmod(a[d], k, 1000000007));
        ans += 1000000007;
        ans = ans % 1000000007;
      }
    }
  }
}
int main() {
  long long int node, x, y, color, tot;
  cin >> node >> k;
  ans = bigmod(node, k, 1000000007);
  for (int i = 0; i < node - 1; i++) {
    cin >> x >> y >> color;
    m[make_pair(x, y)] = color;
    m[make_pair(y, x)] = color;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (z = 0; z < 100004; z++) a[z] = 1;
  dfs(1);
  ans -= (bigmod(a[1], k, 1000000007));
  ans += 1000000007;
  ans = ans % 1000000007;
  cout << ans << endl;
}
