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;
template <typename T>
bool uax(T &a, const T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool uin(T &a, const T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &out, pair<T1, T2> &p) {
  out << p.first << ' ' << p.second;
  return out;
}
mt19937 rnd(1928);
const long long MOD = 1e9 + 7;
long long n, k;
vector<vector<pair<int, bool>>> g;
vector<int> used;
vector<long long> szc;
void dfs(int v) {
  szc.back()++;
  for (int i = 0; i < g[v].size(); i++) {
    if (used[g[v][i].first] == 0 && !g[v][i].second) {
      used[g[v][i].first] = used[v];
      dfs(g[v][i].first);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(15);
  cin >> n >> k;
  g.resize(n);
  used.resize(n);
  int x, y, z;
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y >> z;
    x--;
    y--;
    g[x].push_back(make_pair(y, z));
    g[y].push_back(make_pair(x, z));
  }
  int cnt = 1;
  for (int i = 0; i < n; i++) {
    if (used[i] == 0) {
      szc.push_back(0);
      used[i] = cnt++;
      dfs(i);
    }
  }
  long long ans = 1;
  for (int i = 0; i < k; i++) {
    ans = (ans * n) % MOD;
  }
  for (int i = 0; i < szc.size(); i++) {
    long long x = 1;
    for (int j = 0; j < k; j++) {
      x = (x * szc[i]) % MOD;
    }
    ans = (ans - x + MOD) % MOD;
  }
  cout << ans;
}
