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 = 3e5 + 5, mod = 1e9 + 7;
template <class o>
inline void qr(o &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + (c ^ 48);
    c = getchar();
  }
  x *= f;
}
template <class o>
void qw(o x) {
  if (x / 10) qw(x / 10);
  putchar(x % 10 + 48);
}
template <class o>
void pr1(o x) {
  if (x < 0) putchar('-'), x = -x;
  qw(x);
  putchar(' ');
}
template <class o>
void pr2(o x) {
  if (x < 0) putchar('-'), x = -x;
  qw(x);
  puts("");
}
inline int ksm(int a, int b = mod - 2) {
  int ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) ans = 1ll * ans * a % mod;
  return ans;
}
inline int add(int a, int b) { return a += b, a >= mod ? a - mod : a; }
inline int sub(int a, int b) { return a -= b, a < 0 ? a + mod : a; }
vector<pair<int, int> > e[N];
int siz[N], col[N], cnt, pw[N];
void dfs(int x) {
  ++siz[cnt];
  col[x] = cnt;
  for (pair<int, int> t : e[x]) {
    int y = t.first, z = t.second;
    if (z || col[y]) continue;
    dfs(y);
  }
}
void solve() {
  int n, k;
  qr(n), qr(k);
  for (int i = 1; i <= n; ++i) pw[i] = ksm(i, k);
  for (int i = 1; i < n; ++i) {
    int x, y, z;
    qr(x), qr(y), qr(z);
    e[x].emplace_back(y, z);
    e[y].emplace_back(x, z);
  }
  int ans = pw[n];
  for (int i = 1; i <= n; ++i)
    if (!col[i]) {
      ++cnt;
      dfs(i);
      ans = sub(ans, pw[siz[cnt]]);
    }
  pr2(ans);
}
int main() {
  solve();
  return 0;
}
