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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int N = 100005;
bool vis[N];
vector<pair<int, int> > g[N];
long long c = 0;
long long power(long long x, long long y);
long long powerm(long long x, long long y, int mod2);
void seive(bool arr[], int n);
void dfs(int node);
void solve() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 1; i < n; i++) {
    long long x, y, z;
    cin >> x >> y >> z;
    g[x].push_back({y, z}), g[y].push_back({x, z});
  }
  long long ans = power(n, k);
  memset(vis, false, sizeof(vis));
  for (long long i = 1; i < n + 1; i++) {
    c = 0;
    if (!vis[i]) {
      dfs(i);
      ans = (ans - power(c, k) + 1000000007) % 1000000007;
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  while (t--) solve();
}
void dfs(int node) {
  vis[node] = true;
  c++;
  for (auto x : g[node]) {
    if (!vis[x.first] && x.second == 0) {
      dfs(x.first);
    }
  }
}
long long power(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1)
      ans = ((long long)(ans % 1000000007) * (long long)(x % 1000000007)) %
            1000000007;
    x = ((long long)(x % 1000000007) * (long long)(x % 1000000007)) %
        1000000007;
    y = y >> 1;
  }
  return ans % 1000000007;
}
long long powerm(long long x, long long y, int mod2) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ((long long)(ans % mod2) * (long long)(x % mod2)) % mod2;
    x = ((long long)(x % mod2) * (long long)(x % mod2)) % mod2;
    y = y >> 1;
  }
  return ans % mod2;
}
