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 inf = 0x3f3f3f3f3f3f3f3f;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void zz() { ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0); }
const long long mx = 1e5 + 5;
map<long long, vector<long long> > mp;
long long vis[mx];
long long ans = 0;
long long xx = 0;
long long power(long long a, long long n) {
  if (!n) return 1;
  long long pt = power(a, n / 2);
  pt *= pt, pt %= 1000000007;
  if (n & 1) pt *= a, pt %= 1000000007;
  return pt;
}
void dfs(long long u, long long p = -1) {
  vis[u] = 1;
  for (auto it : mp[u]) {
    if (vis[it]) continue;
    xx += 1;
    dfs(it, u);
  }
  return;
}
int32_t main() {
  zz();
  long long n, k;
  cin >> n >> k;
  long long cnt = 0;
  set<long long> s;
  for (long long i = 0; i < n - 1; i++) {
    long long x, y, z;
    cin >> x >> y >> z;
    if (z == 0) {
      mp[x].emplace_back(y);
      mp[y].emplace_back(x);
    }
    if (z) cnt += 1;
  }
  for (auto it : mp) {
    xx = 0;
    if (!vis[it.first]) {
      dfs(it.first);
      ans += power(xx + 1, k) - xx - 1;
      ans %= 1000000007;
    }
  }
  long long x = power(n, k);
  long long second = 0;
  second = (x - ans - n) % 1000000007;
  cout << second << endl;
}
