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;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void __print(int x) { cout << x; }
void __print(long x) { cout << x; }
void __print(long long x) { cout << x; }
void __print(unsigned x) { cout << x; }
void __print(unsigned long x) { cout << x; }
void __print(unsigned long long x) { cout << x; }
void __print(float x) { cout << x; }
void __print(double x) { cout << x; }
void __print(long double x) { cout << x; }
void __print(char x) { cout << '\'' << x << '\''; }
void __print(const char *x) { cout << '\"' << x << '\"'; }
void __print(const string &x) { cout << '\"' << x << '\"'; }
void __print(bool x) { cout << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cout << '{';
  __print(x.first);
  cout << ',';
  __print(x.second);
  cout << '}';
}
template <typename T>
void __print(const T &x) {
  int first = 0;
  cout << '{';
  for (auto &i : x) cout << (first++ ? "," : ""), __print(i);
  cout << "}";
}
void _print() { cout << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cout << ", ";
  _print(v...);
}
template <class T>
void remdup(vector<T> &v) {
  sort((v).begin(), (v).end());
  v.erase(unique((v).begin(), (v).end()), end(v));
}
template <typename T>
struct cmp {
  bool operator()(const T &p1, const T &p2) {}
};
const long long N = 2e5 + 100;
const long long mod = 1e9 + 7;
vector<long long> a[N];
long long check[N];
long long count1 = 0;
long long power(long long x, long long n) {
  long long result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % mod;
    x = (x * x) % mod;
    n = n / 2;
  }
  return result;
}
void dfs(long long x) {
  count1++;
  check[x] = 1;
  for (auto &it : a[x])
    if (!check[it]) dfs(it);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ;
  long long n, k, x, y, z;
  cin >> n >> k;
  for (long long i = (0); i < (n - 1); ++i) {
    cin >> x >> y >> z;
    if (!z) {
      a[x].push_back(y);
      a[y].push_back(x);
    }
  }
  long long ans = 0;
  for (long long i = (1); i < (n + 1); ++i) {
    if (!check[i]) dfs(i);
    ans += power(count1, k);
    ans = ans % mod;
    count1 = 0;
  }
  cout << (power(n, k) - ans + mod) % mod << "\n";
  return 0;
}
