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;
inline long long max(long long a, long long b) { return ((a > b) ? a : b); }
inline long long min(long long a, long long b) { return ((a > b) ? b : a); }
inline long long gcd(long long a, long long b) {
  if (b == 0) return a;
  a %= b;
  return gcd(b, a);
}
long long power(long long x, long long y) {
  if (y == 0) return 1;
  long long a = power(x, y / 2);
  if (y % 2 == 0)
    return a * a;
  else
    return x * a * a;
}
long long powermod(long long x, long long ex, long long md) {
  long long ans = 1ll;
  while (ex > 0) {
    if (ex & 1ll) ans = (ans * x) % md;
    ex >>= 1ll;
    x = (x * x) % md;
  }
  return ans;
}
const long long inf = 1e18 + 9;
const long long mod = 1e9 + 7;
const long double PI = acos(-1);
const long double eps = 1e-9;
const long long N = 1e6 + 1;
long long n, k;
vector<long long> red[N];
bool vis[N];
set<long long> second;
void dfs(long long node) {
  vis[node] = 1;
  second.insert(node);
  for (auto child : red[node]) {
    if (!vis[child]) {
      dfs(child);
    }
  }
}
void solve() {
  cin >> n >> k;
  for (int i = 0; i < n - 1; i++) {
    long long x, y, c;
    cin >> x >> y >> c;
    if (c == 0) {
      red[x].push_back(y);
      red[y].push_back(x);
    }
  }
  long long ans = 0;
  ans = powermod(n, k, mod);
  long long x = 0;
  for (long long i = 1; i <= n; i++) {
    if (!vis[i]) {
      second.clear();
      dfs(i);
      long long len = second.size();
      x += powermod(len, k, mod);
      x %= mod;
    }
  }
  cout << (ans - x + mod) % mod << "\n";
}
int main() {
  {
    ios ::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  };
  cout << fixed << setprecision(20);
  long long NTC = 1;
  long long PTC = 0;
  while (PTC++ < NTC) {
    solve();
  }
  cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC
       << "ms\n";
}
