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>
template <typename T>
inline void read(T &x) {
  char c;
  bool nega = 0;
  while ((!isdigit(c = getchar())) && (c != '-'))
    ;
  if (c == '-') {
    nega = 1;
    c = getchar();
  }
  x = c - 48;
  while (isdigit(c = getchar())) x = x * 10 + c - 48;
  if (nega) x = -x;
}
template <typename T>
inline void writep(T x) {
  if (x > 9) writep(x / 10);
  putchar(x % 10 + 48);
}
template <typename T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  writep(x);
  putchar(' ');
}
template <typename T>
inline void writeln(T x) {
  write(x);
  putchar('\n');
}
using namespace std;
const int mod = 1000000007;
int n, k, sz[100005], par[100005], ans;
pair<pair<int, int>, int> ed[100005];
vector<int> a[100005];
void Dfs(int u) {
  sz[u] = 1;
  for (int v : a[u]) {
    if (v == par[u]) continue;
    par[v] = u;
    Dfs(v);
    sz[u] += sz[v];
  }
}
int Pow(int a, int b) {
  int res = 1;
  for (; b; a = (1ll * a * a) % mod, b >>= 1)
    if (b & 1) res = (1ll * res * a) % mod;
  return res;
}
int Mul(int x, int y) { return (1ll * x * y) % mod; }
void Add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
void Sub(int &x, int y) {
  x -= y;
  if (x < 0) x += mod;
}
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> k;
  for (int i = (1); i <= (n - 1); ++i) {
    int x, y, z;
    cin >> x >> y >> z;
    ed[i] = {{x, y}, z};
    if (z) continue;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  ans = Pow(n, k);
  for (int i = (1); i <= (n); ++i) {
    if (sz[i]) continue;
    Dfs(i);
    Sub(ans, Pow(sz[i], k));
  }
  cout << ans;
  return 0;
}
