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;
long long GCD(long long x, long long y) {
  if (y == 0) return x;
  return GCD(y, x % y);
}
long long LCM(long long x, long long y) { return (x * y) / (GCD(x, y)); }
long long LOGK(long long x, long long k) {
  if (x >= k) return 1 + LOGK(x / k, k);
  return 0;
}
long long MPOW(long long a, long long b, long long m) {
  if (b == 0) return 1;
  long long x = MPOW(a, b / 2, m);
  x = (x * x) % m;
  if (b % 2 == 1) x = (x * a) % m;
  return x;
}
long long MINV(long long a, long long m) { return MPOW(a, m - 2, m); }
class pnc {
  long long FACT_MAX, MODU;
  vector<long long> fact;

 public:
  pnc(long long n, long long m) {
    FACT_MAX = n;
    fact.resize(FACT_MAX);
    MODU = m;
    MFACT_INIT(MODU);
  }
  void MFACT_INIT(long long m) {
    fact[0] = 1;
    for (long long i = 1; i < FACT_MAX; ++i) fact[i] = (i * fact[i - 1]) % MODU;
  }
  long long MFACT(long long n) { return fact[n]; }
  long long PERM(long long n, long long r) {
    return (fact[n] * ::MINV(fact[n - r], MODU)) % MODU;
  }
  long long COMB(long long n, long long r) {
    return (PERM(n, r) * ::MINV(fact[r], MODU)) % MODU;
  }
};
long long n, k;
vector<pair<long long, long long> > adj[200005];
bool visited[200005] = {false};
long long tans = 0;
void dfs(long long node) {
  visited[node] = true;
  tans++;
  for (auto i : adj[node]) {
    if (!visited[i.first] && !i.second) {
      dfs(i.first);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long test = 1;
  while (test--) {
    cin >> n >> k;
    long long x, y, color;
    for (long long i = 1; i < n; i++) {
      cin >> x >> y >> color;
      adj[x].push_back({y, color});
      adj[y].push_back({x, color});
    }
    long long ans = MPOW(n, k, 1000000007);
    for (long long i = 1; i <= n; i++) {
      tans = 0;
      if (!visited[i]) {
        dfs(i);
      }
      long long temp = MPOW(tans, k, 1000000007);
      ans -= temp - 1000000007;
      ans %= 1000000007;
    }
    cout << ans;
  }
}
