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;
const long double PI = 4 * atan((long double)1);
long long int get_hash(string s) {
  long long int N = 1000001;
  long long int base[N], A = 11, MD = 1110111110111;
  base[0] = 1;
  for (int i = (1); i < (N); ++i) base[i] = (base[i - 1] * A) % MD;
  long long int hs = 0;
  for (int i = (0); i < (s.size()); ++i) {
    hs += (s[i] * base[i]);
    hs %= MD;
  }
  return hs;
}
long long power(long long a, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res = (res % MOD * a % MOD) % MOD;
    a = (a % MOD * a % MOD) % MOD;
    n /= 2;
  }
  return res % MOD;
}
int par(int i, int* dp) {
  if (i == dp[i]) return i;
  return (par(dp[i], dp));
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, k;
  cin >> n >> k;
  int dp[n + 1];
  for (int i = (0); i < (n + 1); ++i) dp[i] = i;
  for (int i = (0); i < (n - 1); ++i) {
    int x, y, c;
    cin >> x >> y >> c;
    if (c == 0) dp[par(x, dp)] = dp[par(y, dp)] = min(par(x, dp), par(y, dp));
  }
  map<int, int> mpa;
  map<int, int>::iterator it;
  for (int i = (1); i < (n + 1); ++i) mpa[par(i, dp)]++;
  long long int ans = 0;
  for (it = mpa.begin(); it != mpa.end(); ++it) {
    int sp = it->second;
    long long int x = power(sp, k) % MOD;
    ans = (ans % MOD + x % MOD) % MOD;
  }
  long long int tot = power(n, k) % MOD;
  cout << (tot - ans + MOD) % MOD;
  return 0;
}
