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 mod = 1e9 + 7, N = 1e5 + 10;
vector<pair<long long, long long> > adj[N];
bool visited[N];
long long co[N];
long long n, k;
long long sum = 0;
long long bin_expo(long long x, long long y, long long m) {
  long long res = 1;
  x = x % m;
  while (y > 0) {
    if (y & 1) res = (res * x) % m;
    y = y / 2;
    x = (x * x) % m;
  }
  return res;
}
void dfs(long long s, long long p) {
  visited[s] = true;
  for (auto x : adj[s]) {
    if (visited[x.first] == false) {
      if (x.second == 0) {
        co[p]++;
        dfs(x.first, p);
      } else {
        co[x.first]++;
        dfs(x.first, x.first);
      }
    }
  }
}
void solve() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    co[i] = 0;
    visited[i] = false;
  }
  for (int i = 0; i < n - 1; i++) {
    long long x, y, cr;
    cin >> x >> y >> cr;
    x--;
    y--;
    adj[x].push_back(make_pair(y, cr));
    adj[y].push_back(make_pair(x, cr));
  }
  co[0]++;
  dfs(0, 0);
  long long ans = bin_expo(n, k, mod) % mod - n % mod;
  for (int i = 0; i < n; i++) {
    if (co[i] != 0) {
      sum = (sum + bin_expo(co[i], k, mod) - co[i]) % mod;
    }
  }
  cout << ans - sum << "\n";
  return;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int test = 1;
  while (test--) {
    solve();
  }
}
