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 long long MOD = 1000000007;
int findchar(string s, char c) {
  int i = 0;
  while (i < s.length()) {
    if (s[i] == c) return i;
    ++i;
  }
  return -1;
}
void shift(vector<int> &ar, int l, int r) {
  int x = ar[r];
  for (int i = l; i < r; i++) {
    ar[i + 1] = ar[i];
  }
  ar[l] = x;
  for (int i = l; i < r; i++) {
    cout << i << " th element : " << ar[i] << "\n";
  }
  cout << "\n";
}
long long isPrime(long long n) {
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return i;
  }
  return -1;
}
int greatestdivisor(int n, int k) {
  int maxi = n;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      maxi = min(maxi, n / i);
    }
  }
  return maxi;
}
int binary(vector<int> ar, int n, int item) {
  int beg = 0, last = n - 1;
  int mid;
  while (beg <= last) {
    mid = (beg + last) / 2;
    if (item == ar[mid])
      return mid;
    else if (item < ar[mid])
      last = mid - 1;
    else
      beg = mid + 1;
  }
  return last;
}
long long gcd(long long a, long long b) {
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
bool isOverflow(long long a, long long b) {
  if (a == 0 || b == 0) return false;
  long long result = a * b;
  if (a == result / b)
    return false;
  else
    return true;
}
bool cmp(const pair<long long, long long> &a,
         const pair<long long, long long> &b) {
  return a.second - a.first > b.second - b.first;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<pair<long long, long long> > ar;
  vector<long long> color(n - 1);
  long long ans = pow(n, k);
  ans %= MOD;
  vector<bool> vis(n + 1, false);
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    ar.push_back(make_pair(x, y));
    cin >> color[i];
    if (color[i] == 0) {
      vis[x] = true;
      vis[y] = true;
    }
  }
  long long black = 0;
  for (long long i = 1; i <= n; i++) {
    if (!vis[i]) {
      black++;
    }
  }
  ans -= black;
  ans %= MOD;
  ans -= pow(n - black, k);
  ans %= MOD;
  cout << ans << "\n";
  return 0;
}
