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 double pi = acos(-1);
const int N = 1e5 + 7, mod = 1000000007, M = 2e9;
const long long MAX = 5e18;
int moding(long long x) {
  while (x < 0) x += mod;
  if (x >= mod) x %= mod;
  return x;
}
int n, k, u, v, x, c[N];
long long ans = 1;
struct dsu {
  int sz[N], cl[N];
  void init() {
    fill(sz, sz + N, 0);
    iota(cl, cl + N, 0);
  }
  int par(int x) { return cl[x] = (x == cl[x] ? x : par(cl[x])); }
  void con(int x, int y) {
    x = par(x);
    y = par(y);
    if (x == y) return;
    if (sz[y] > sz[x]) swap(x, y);
    sz[x] += sz[y];
    cl[y] = x;
  }
} tree;
long long fac(int x) {
  long long ret = 1;
  for (int i = 0; i < k; i++) ret = moding(ret * x);
  return ret;
}
int main() {
  scanf("%d%d", &n, &k);
  tree.init();
  for (int i = 1; i < n; i++) {
    scanf("%d%d%d", &u, &v, &x);
    if (!x) tree.con(u, v);
  }
  for (int i = 1; i <= n; i++) c[tree.par(i)]++;
  ans = fac(n);
  for (int i = 1; i <= n; i++) ans = moding(ans - fac(c[i]));
  cout << ans;
  return 0;
}
