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 parent[100000 + 5];
long long Size[100000 + 5];
long long power(long long x, 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;
}
void first(long long n) {
  for (long long i = 1; i <= n; i++) {
    parent[i] = i;
    Size[i] = 1;
  }
}
long long findroot(long long p) {
  while (parent[p] != p) {
    p = parent[p];
  }
  return p;
}
int main() {
  long long n, k, x, y, t, cnt1 = 0, root1, root2, ans = 0, finans, cnt2 = 0;
  long long mod = 1000000007;
  map<long long, long long> m1;
  map<long long, long long> m0;
  scanf("%lld", &n);
  ;
  scanf("%lld", &k);
  ;
  first(n);
  finans = power(n, k, mod);
  for (long long i = 0; i < n - 1; i++) {
    scanf("%lld", &x);
    ;
    scanf("%lld", &y);
    ;
    scanf("%lld", &t);
    ;
    if (t == 1) {
      m1[x] = 1;
      m1[y] = 1;
    }
    if (t == 0) {
      m0[x] = 1;
      m0[y] = 1;
      root1 = findroot(x);
      root2 = findroot(y);
      if (root1 != root2) {
        if (Size[root1] > Size[root2] || Size[root1] == Size[root2]) {
          parent[root2] = root1;
          Size[root1] += Size[root2];
        } else {
          parent[root1] = root2;
          Size[root2] += Size[root1];
        }
      }
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (m1[i] == 1 && m0.find(i) == m0.end()) {
      cnt1++;
    }
    if (m1[i] == 1) {
      cnt2++;
    }
  }
  for (long long j = 1; j <= n; j++) {
    if (m0[j] == 1 && parent[j] == j) {
      long long val = Size[j];
      ans = ((ans % mod) + power(val, k, mod)) % mod;
    }
  }
  if (cnt2 == 0) {
    printf("%s", "0");
    ;
    printf("\n");
    ;
  } else {
    finans = (((finans - ans) % mod) + mod) % mod;
    finans = (((finans - cnt1) % mod) + mod) % mod;
    printf("%lld", finans);
    ;
    printf("\n");
    ;
  }
}
