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 = 1e5 + 5, mod = 1000000007;
struct RBTree {
  int to, Next;
} a[MAXN * 2];
int first[MAXN], vis[MAXN];
int n, k, ans, cnt, newp;
void dfs(int x) {
  vis[x] = 1;
  cnt++;
  for (int i = first[x]; i; i = a[i].Next) {
    int To = a[i].to;
    if (vis[To]) continue;
    dfs(To);
  }
}
int Qkpow(int b, int p) {
  int s = 1;
  while (p) {
    if (p & 1) s = s * b % mod;
    b = b * b % mod;
    p >>= 1;
  }
  return s;
}
inline void add(int x, int y) {
  a[++newp] = (RBTree){y, first[x]};
  first[x] = newp;
}
inline int read(void) {
  int num = 0, f = 1;
  char ch;
  while (!isdigit(ch = getchar()))
    if (ch == '-') f = -1;
  while (isdigit(ch)) num = num * 10 + ch - '0', ch = getchar();
  return num * f;
}
int main() {
  n = read(), k = read();
  for (int i = 1; i < n; i++) {
    int u = read(), v = read(), o = read();
    if (!o) add(u, v), add(v, u);
  }
  ans = Qkpow(n, k);
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      cnt = 0;
      dfs(i);
      ans = (ans - Qkpow(cnt, k) + mod) % mod;
    }
  }
  printf("%d\n", ans);
  return 0;
}
