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;
int readInt() {
  int x;
  if (scanf(" %d", &x) != EOF) return x;
  return -1;
}
long long int readLint() {
  long long int x;
  if (cin >> x) return x;
  return -1;
}
string readString() {
  string s;
  if (cin >> s) return s;
  return "";
}
struct UnionFind {
  int n;
  vector<int> parent;
  vector<int> size;
  UnionFind(int n) : n(n) {
    parent = vector<int>(n, -1);
    for (int i = 0; i < n; ++i) parent[i] = i;
    size = vector<int>(n, 1);
  }
  int findset(int u) {
    if (parent[u] == u) return u;
    int x = u;
    while (x != parent[x]) {
      x = parent[x];
    }
    while (u != x) {
      int old = parent[u];
      parent[u] = x;
      u = old;
    }
    return x;
  }
  void attach(int u, int v) {
    parent[u] = v;
    size[v] += size[u];
  }
  void merge(int u, int v) {
    int U = findset(u);
    int V = findset(v);
    if (rand() % 2) {
      attach(U, V);
    } else
      attach(V, U);
  }
  int getComponentSize(int u) { return size[findset(u)]; }
};
const long long int mod = 1000000007;
struct SingleCase {
  bool solveCase() {
    int n = readInt();
    if (n < 0) return false;
    int k = readInt();
    UnionFind zero(n);
    for (int i = 0; i < (n - 1); ++i) {
      int u = readInt() - 1;
      int v = readInt() - 1;
      int c = readInt();
      if (c == 0) {
        zero.merge(u, v);
      }
    }
    vector<long long int> p(k + 1, 0);
    p[0] = 1;
    for (int i = 1; i < p.size(); i++) {
      p[i] = p[i - 1] * n % mod;
    }
    long long int ret = 0;
    for (int i = 0; i < n; ++i) {
      long long int cur = 0;
      long long int size = zero.getComponentSize(i);
      long long int seq = 1;
      for (long long int seq = 1, r = 2; r <= k; r++) {
        cur = (cur + seq * (n - size) % mod * p[k - r] % mod) % mod;
        seq = seq * size % mod;
      }
      ret = (ret + cur) % mod;
    }
    cout << ret << endl;
    return true;
  }
};
int main() {
  while (SingleCase().solveCase()) {
  }
}
