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;
const int maxn = 1e5 + 100;
const int mod = 1e9 + 7;
int cnt[maxn];
vector<pair<int, int> > edge[maxn];
vector<int> q[maxn];
void dfs(int u, int col) {
  cnt[u] = col;
  for (int i = 0; i < edge[u].size(); i++) {
    int to = edge[u][i].first;
    if (cnt[to] || edge[u][i].second) continue;
    dfs(to, col);
  }
}
int main() {
  int n, m, tot = 0;
  scanf("%d %d", &n, &m);
  int st, ed, col;
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d %d %d", &st, &ed, &col);
    edge[st].push_back(make_pair(ed, col));
    edge[ed].push_back(make_pair(st, col));
  }
  for (int i = 1; i <= n; i++) {
    if (!cnt[i]) {
      tot++;
      dfs(i, tot);
    }
  }
  for (int i = 1; i <= n; i++) {
    q[cnt[i]].push_back(i);
  }
  long long sum = 0;
  long long tmp = 1;
  for (int i = 1; i <= m; i++) {
    tmp = tmp * n % mod;
  }
  sum = tmp;
  for (int i = 1; i <= tot; i++) {
    long long tmp = 1;
    for (int j = 0; j < m; j++) {
      tmp = tmp * q[i].size() % mod;
    }
    sum = (sum - tmp + mod) % mod;
  }
  printf("%lld\n", sum);
}
