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 long long infl = 1e18 + 5;
long long n, m, k, q, x, y, f, val, t, i, j;
long long ind, cnt, sz, sm, mx, mn, ans;
long long a[1000004], vis[1000004];
vector<long long> v[1000004];
map<pair<long long, long long>, long long> mp;
long long pow_mod(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res *= a;
      res %= 1000000007;
    }
    a *= a;
    a %= 1000000007;
    b /= 2;
  }
  return res;
}
long long dfs(int i) {
  if (vis[i]) return 0;
  vis[i] = 1;
  long long re = 1;
  for (auto it : v[i]) {
    if (mp[{it, i}] == 0) {
      re += dfs(it);
    }
  }
  return re;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (fopen("inp.txt", "r")) {
    freopen("myfile.txt", "w", stdout);
    freopen("inp.txt", "r", stdin);
  }
  cin >> n >> k;
  for (i = 0; i < n - 1; i++) {
    cin >> x >> y >> val;
    v[x].emplace_back(y);
    v[y].emplace_back(x);
    mp[{x, y}] = val;
    mp[{y, x}] = val;
  }
  ans = pow_mod(n, k);
  for (i = 1; i < n + 1; i++) {
    if (!vis[i]) {
      sz = dfs(i);
      ans -= pow_mod(sz, k);
      ans += 1000000007 + 1000000007;
      ans %= 1000000007;
    }
  }
  cout << ans;
  return 0;
}
