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 = 2e5 + 10, mo = 1e9 + 7;
int n, k;
int ds[N], si[N];
int dsu(int x) { return x == ds[x] ? ds[x] : dsu(ds[x]); }
long long mul(int a, int n) {
  if (n == 0) return 1;
  long long t = mul(a, n / 2);
  t = t * t % mo;
  return (n & 1) ? t * a % mo : t;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1, _end = n; i <= _end; i++) si[i] = 1, ds[i] = i;
  for (int i = 1, _end = n - 1; i <= _end; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    if (z == 0) {
      int X = dsu(x), Y = dsu(y);
      ds[X] = Y;
      si[Y] += si[X];
    }
  }
  long long ans = 0;
  for (int i = 1, _end = n; i <= _end; i++)
    if (ds[i] == i) {
      ans += mul(si[i], k);
      ans %= mo;
    }
  long long tot = mul(n, k);
  printf("%lld", tot - ans);
}
