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 maxn = 1e5 + 10;
const long long mod = 1e9 + 7;
long long n, k, num = 1, cnt;
vector<long long> v[maxn];
bool vis[maxn];
long long color[maxn];
map<long long, long long> mp;
void bfs(long long x) {
  queue<long long> q;
  vis[x] = true;
  ++cnt;
  q.push(x);
  while (!q.empty()) {
    long long kx = q.front();
    q.pop();
    for (long long i = 0; i < v[kx].size(); ++i) {
      long long k = v[kx][i];
      if (!vis[k]) {
        vis[k] = true;
        q.push(k);
        ++cnt;
      }
    }
  }
}
long long ksm(long long a, long long x) {
  if (x == 0) return 1;
  if (x == 1) return a % mod;
  long long tmp = ksm(a, x / 2) % mod;
  if (x % 2 == 0)
    return (tmp * tmp) % mod;
  else
    return (((tmp * tmp) % mod) * a) % mod;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  long long ans = ksm(n, k);
  for (long long i = 1; i < n; ++i) {
    long long a, b, c;
    cin >> a >> b >> c;
    if (c == 0) {
      v[a].push_back(b);
      v[b].push_back(a);
    }
  }
  for (long long i = 1; i <= n; ++i) {
    if (!vis[i]) {
      cnt = 0;
      bfs(i);
      long long BDFCL = ksm(cnt, k);
      ans -= BDFCL;
      if (ans < 0) ans += mod;
    }
  }
  cout << ans % mod << endl;
  return 0;
}
