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 maxn = 100001;
int n, k;
int tot = 0;
int head[maxn];
struct abcd {
  int to, nxt;
  long long f;
} table[2 * maxn];
void Add(int x, int y, long long z) {
  table[++tot].to = y;
  table[tot].f = z;
  table[tot].nxt = head[x];
  head[x] = tot;
}
int flag[maxn];
namespace Union_Find_Set {
int fa[maxn], size[maxn], max_val[maxn], min_val[maxn], vis[maxn];
int Find(int x) {
  if (!fa[x]) size[x] = 1, fa[x] = x;
  if (fa[x] == x) return x;
  return fa[x] = Find(fa[x]);
}
void Union(int x, int y) {
  x = Find(x);
  y = Find(y);
  if (x == y) return;
  fa[x] = y;
  size[y] += size[x];
}
}  // namespace Union_Find_Set
long long kpow(long long x, long long n, long long p) {
  long long s = 1;
  while (n) {
    if (n & 1) s = (s * x) % p;
    x = (x * x) % p;
    n = n >> 1;
  }
  return s;
}
void dfs(int x, int p) {
  flag[x] = 1;
  using namespace Union_Find_Set;
  for (int i = head[x]; i; i = table[i].nxt) {
    if (table[i].to == p || table[i].f == 1) continue;
    Union(x, table[i].to);
    dfs(table[i].to, x);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  long long ans = kpow(n, k, 1000000007);
  using namespace Union_Find_Set;
  for (int i = 0; i < n - 1; i++) {
    int x, y, w;
    cin >> x >> y >> w;
    Add(x, y, w);
    Add(y, x, w);
  }
  for (int i = 1; i <= n; i++) {
    if (!flag[i]) dfs(i, -1);
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[Find(i)]) {
      long long t = kpow(size[Find(i)], k, 1000000007);
      ans = (ans - t + 1000000007) % 1000000007;
      vis[Find(i)] = 1;
    }
  }
  cout << ans << endl;
}
