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;
bool cmp(vector<long long int> v1, vector<long long int> v2) {
  return (v1.size() < v2.size());
}
int find(long long int parent[], long long int i) {
  if (parent[i] == -1) return i;
  return find(parent, parent[i]);
}
void Union(long long int parent[], long long int x, long long int y) {
  long long int xset = find(parent, x);
  long long int yset = find(parent, y);
  if (xset != yset) {
    parent[xset] = yset;
  }
}
long long int maxSubArraySum(long long int a[], long long int size) {
  long long int max_so_far = INT_MIN, max_ending_here = 0;
  for (long long int i = 0; i < size; i++) {
    max_ending_here = max_ending_here + a[i];
    if (max_so_far < max_ending_here) max_so_far = max_ending_here;
    if (max_ending_here < 0) max_ending_here = 0;
  }
  return max_so_far;
}
long long int gcdExtended(long long int a, long long int b, long long int *x,
                          long long int *y) {
  if (a == 0) {
    *x = 0;
    *y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long int power(long long int x, long long int n) {
  int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % 1000000007;
    x = (x * x) % 1000000007;
    n = n / 2;
  }
  return result;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0)
    return b;
  else
    return gcd(b % a, a);
}
vector<long long int> tree;
vector<long long int> v[100001];
long long int st[100001], e[100001], tim = 0, mar[100001] = {0};
bool visit[100001] = {false};
long long int coun = 0;
long long int dfs(long long int root) {
  visit[root] = true;
  coun++;
  long long int i;
  tree.push_back(root);
  for (i = 0; i < v[root].size(); i++) {
    if (visit[v[root][i]] == false) {
      dfs(v[root][i]);
    }
  }
  return coun;
}
int main() {
  long long int i, j, k, n, m, y, z, count[100001] = {0}, x = 0, ans = 0;
  scanf("%lld", &n);
  scanf("%lld", &k);
  for (i = 0; i < n - 1; i++) {
    scanf("%lld", &x);
    scanf("%lld", &y);
    scanf("%lld", &z);
    if (z == 0) v[x].push_back(y);
    v[y].push_back(x);
  }
  x = 0;
  count[x++] = dfs(1);
  for (i = 1; i < n + 1; i++) {
    coun = 0;
    if (visit[i] == false) count[x++] = dfs(i);
  }
  m = power(n, k);
  cout << m << " ";
  for (i = 0; i < x; i++) {
    y = power(count[i], k);
    m = (m - y + 1000000007) % 1000000007;
  }
  cout << m;
  return 0;
}
