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. 

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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;
vector<list<long long>> gh;
vector<long long> count;
int const MOD = 1e9 + 7;
int visited[100000];
vector<long long> components(100000, 0);
int pow_m(int x, int y) {
  if (y == 0) {
    return 1;
  }
  long long u = pow_m(x, y / 2);
  u = u * u % MOD;
  if (y % 2 == 1) {
    u = u * x % MOD;
  }
  return u;
}
int dfs(vector<list<long long>>& g, long long u, long long startnode) {
  if (components[startnode] == 0) components[startnode]++;
  int ret = 1;
  for (auto it = gh[u].begin(); it != gh[u].end(); it++) {
    if (visited[*it] || startnode == *it) continue;
    visited[*it] = 1;
    components[startnode]++;
    ret += dfs(g, *it, startnode);
  }
  return ret;
}
int main() {
  long long n, count, k;
  cin >> n >> k;
  long long t = n;
  long long a, b, c;
  for (int i = 0; i < 100000; i++) {
    list<long long> l;
    gh.push_back(l);
    visited[i] = 0;
    gh[i].push_back(i);
    components[i] = 0;
  }
  for (int i = 1; i < n; i++) {
    cin >> a >> b >> c;
    if (c == 0) {
      gh[a].push_back(b);
      gh[b].push_back(a);
    }
  }
  long long red = 0;
  for (int node = 1; node <= n; node++) {
    if (!visited[node]) red += (pow_m(dfs(gh, node, node), k) + MOD) % MOD;
  }
  long long total = pow_m(t, k);
  long long result = total - red;
  result = result % (1000000007);
  cout << result << endl;
  return 0;
}
