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 N = ((int)1e5) + 5;
const int M = ((int)1e3) + 5;
long long int n, k;
long long int posetler;
long long int dad[N];
long long int size[N];
bool vis[N];
vector<long long int> v;
long long int find(long long int a) {
  if (dad[a] == a) return a;
  return dad[a] = find(dad[a]);
}
void unite(long long int a, long long int b) {
  size[find(b)] += size[find(a)];
  dad[find(a)] = find(b);
}
long long int carp(long long int a, long long int b) {
  return ((a % ((int)(1e9) + 7)) * (b % ((int)(1e9) + 7))) % ((int)(1e9) + 7);
}
long long int fastPow(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  if (b % 2 == 1) {
    long long int x = fastPow(a, b / 2) % ((int)(1e9) + 7);
    return carp(x, carp(x, a)) % ((int)(1e9) + 7);
  }
  long long int x = fastPow(a, b / 2) % ((int)(1e9) + 7);
  return carp(x, x) % ((int)(1e9) + 7);
}
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (long long int i = 1; i <= n; i++) dad[i] = i, size[i] = 1;
  for (long long int i = 1; i <= n - 1; i++) {
    long long int a, b, c;
    cin >> a >> b >> c;
    if (!c) {
      unite(a, b);
      v.push_back(a);
      v.push_back(b);
    }
  }
  for (auto i : v) {
    if (!vis[find(i)]) {
      vis[find(i)] = true;
      posetler += fastPow(size[find(i)], k);
      posetler %= ((int)(1e9) + 7);
    }
  }
  long long int party = 0;
  for (long long int i = 1; i <= n; i++) {
    if (size[find(i)] == 1) party++;
  }
  posetler += party;
  long long int ans =
      (fastPow(n, k) - (posetler % ((int)(1e9) + 7))) % ((int)(1e9) + 7);
  cout << ans << "\n";
}
