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 gcdExtended(int a, int b, int *x, int *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  int x1, y1;
  int gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long int modInverse(int a, int m) {
  int x, y;
  int g = gcdExtended(a, m, &x, &y);
  long long int res = ((long long int)x % m + (long long int)m) % m;
  return res;
}
long long int poww(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return poww(x * x, n / 2);
  else
    return x * poww(x * x, (n - 1) / 2);
}
long long int modularExponentiation(long long int x, long long int n,
                                    long long int M) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = (result * x) % M;
    x = (x * x) % M;
    n = n / 2;
  }
  return result;
}
inline long long int max(long long int a, long long int b) {
  if (a > b)
    return a;
  else
    return b;
}
inline long long int min(long long int a, long long int b) {
  if (a > b)
    return b;
  else
    return a;
}
int gcd(int a, int b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a == b) return a;
  if (a > b)
    return gcd(a - b, b);
  else
    return gcd(a, b - a);
}
long long int ncr(int n, int r) {
  long long int ans = 1;
  r = min(r, n - r);
  for (int i = 0; i < r; i++) {
    ans = (ans * (n - i) / (i + 1)) % 1000000007;
  }
  return ans;
}
bool vis[100001] = {0};
set<pair<int, int> > s;
vector<int> v[100001];
void dfs(int start, long long int &z) {
  if (vis[start]) return;
  vis[start] = 1;
  for (auto x : v[start]) {
    if (!vis[x] && s.find({x, start}) == s.end() &&
        s.find({start, x}) == s.end()) {
      z++;
      dfs(x, z);
    }
  }
  return;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    long long int n, m, k;
    cin >> n >> k;
    vector<int> req;
    for (int i = 0; i < n - 1; i++) {
      int a, b, c;
      cin >> a >> b >> c;
      if (a > b) swap(a, b);
      if (c == 1) {
        s.insert(make_pair(a, b));
        req.push_back(a);
        req.push_back(b);
      }
      v[a].push_back(b);
      v[b].push_back(a);
    }
    long long int ans = modularExponentiation(n, k, 1000000007);
    if (req.size() == 0) {
      cout << "0";
      return 0;
    }
    for (auto x : req) {
      long long int z = 1;
      if (!vis[x]) {
        dfs(x, z);
        ans = ans - modularExponentiation(z, k, 1000000007);
        ans += (long long int)100 * 1000000007;
        ans %= 1000000007;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}
