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 target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long INF = 1e18;
const long long MOD = 1e9 + 7;
long long logn(long long n) {
  long long x = long long(log2(long double(n)));
  if ((1ll << (x)) != n) ++x;
  return x;
}
vector<long long> used;
vector<vector<long long>> g;
void dfs(long long v, long long color, long long p = -1) {
  used[v] = color;
  for (auto to : g[v]) {
    if (to == p) continue;
    dfs(to, color, v);
  }
}
long long f(long long n, long long p) {
  if (p == 0) return 1;
  return (f(n, p - 1) * n) % MOD;
}
inline int source() {
  long long n, k;
  cin >> n >> k;
  g.resize(n);
  for (int i = 1; i < n; ++i) {
    long long v, to, color;
    cin >> v >> to >> color;
    --v, --to;
    if (color == 0) {
      g[v].push_back(to);
      g[to].push_back(v);
    }
  }
  long long kk = 0;
  used.assign(n, 0);
  for (int i = 0; i < n; ++i) {
    if (used[i]) continue;
    ++kk;
    dfs(i, kk);
  }
  vector<long long> comp(k, 0);
  for (int i = 0; i < n; ++i) comp[used[i] - 1]++;
  long long ans = 0;
  if (kk == 1) {
    cout << 0;
    return 0;
  }
  ans = f(n, k);
  for (int i = 0; i < kk; ++i) {
    ans = (ans + MOD - f(comp[i], k)) % MOD;
  }
  cout << ans;
  return 0;
}
int main() {
  srand(time(NULL));
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  source();
  return 0;
}
