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;
int n, k;
long long ans = 0;
vector<pair<int, bool> > son[100005];
namespace Union {
int father[100005];
int vertex[100005];
void init() {
  int i, j;
  for (i = 1; i <= n; i++) {
    father[i] = i;
    vertex[i] = 1;
  }
}
int find(int x) {
  int i, j;
  if (father[x] == x) return x;
  return father[x] = find(father[x]);
}
bool isConnect(int x, int y) { return find(x) == find(y); }
void connect(int x, int y) {
  vertex[find(y)] += vertex[find(x)];
  father[find(x)] = find(y);
}
}  // namespace Union
long long quickPow(int x, int y, int mod) {
  int i, j;
  if (y == 0) return 1;
  long long subPro = quickPow(x, y / 2, mod) % mod;
  if (y % 2 == 0) {
    return subPro * subPro % mod;
  } else {
    return subPro * subPro * x % mod;
  }
}
int main() {
  int i, j;
  cin >> n >> k;
  Union::init();
  for (i = 0; i < n - 1; i++) {
    int x, y;
    bool z;
    cin >> x >> y >> z;
    son[x].push_back(make_pair(y, z));
    son[y].push_back(make_pair(x, z));
  }
  ans = quickPow(n, k, 1000000007);
  for (i = 1; i <= n; i++) {
    for (j = 0; j < son[i].size(); j++) {
      if (!son[i][j].second)
        if (!Union::isConnect(i, (son[i][j].first)))
          Union::connect((son[i][j].first), i);
    }
  }
  for (i = 1; i <= n; i++) {
    if (Union::father[i] == i)
      if (ans > 0) {
        ans -= quickPow((Union::vertex[i]), k, 1000000007);
      }
  }
  cout << ans << endl;
  return 0;
}
