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 power(long long x, unsigned long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long root(long long x, vector<long long> &arr) {
  while (arr[x] != x) {
    arr[x] = arr[arr[x]];
    x = arr[x];
  }
  return x;
}
bool find(long long x, long long y, vector<long long> &arr) {
  if (root(x, arr) == root(y, arr))
    return true;
  else
    return false;
}
void uni(long long x, long long y, vector<long long> &arr,
         vector<long long> &si) {
  long long rx = root(x, arr);
  long long ry = root(y, arr);
  if (si[rx] <= si[ry]) {
    arr[rx] = ry;
    si[ry] += si[rx];
  } else {
    arr[ry] = rx;
    si[rx] += si[ry];
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    long long n, k, ans;
    cin >> n >> k;
    ans = power(n, k, 1000000007);
    if (ans < 0) ans += 1000000007;
    vector<long long> arr(n + 1), si(n + 1), col(n + 1, -1);
    for (long long i = 0; i <= n; i++) {
      arr[i] = i;
      si[i] = 1;
    }
    vector<pair<long long, long long> > v(n - 1);
    for (long long i = 0; i < n - 1; i++) {
      long long x, y, c;
      cin >> x >> y >> c;
      if (c == 0) {
        if (col[x] != 0) col[x] = c;
        if (col[y] != 0) col[y] = c;
        if (!find(x, y, arr)) uni(x, y, arr, si);
      }
    }
    for (long long i = 1; i <= n; i++) {
      if (arr[i] == i) {
        long long x = power(si[i], k, 1000000007);
        if (x < 0) x += 1000000007;
        ans -= x;
        if (ans < 0) ans += 1000000007;
      }
    }
    if (ans < 0) ans += 1000000007;
    cout << ans;
  }
}
