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;
template <typename T>
inline T GCD(T a, T b) {
  a = abs(a);
  b = abs(b);
  if (a < b) swap(a, b);
  while (b) {
    a = a % b;
    swap(a, b);
  }
  return a;
}
template <typename T>
inline T EGCD(T a, T b, T &x, T &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  T x1, y1;
  T d = EGCD(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
template <typename T>
inline T LCM(T x, T y) {
  T tp = GCD(x, y);
  if ((x / tp) * 1. * y > 9e18) return 9e18;
  return (x / tp) * y;
}
template <typename T>
inline T BigMod(T A, T B) {
  T ret = 1;
  while (B) {
    if (B & 1) ret = (ret * A) % 1000000007;
    A = (A * A) % 1000000007;
    B = B >> 1;
  }
  return ret;
}
template <typename T>
inline T InvMod(T A, T M = 1000000007) {
  return BigMod(A, M - 2);
}
vector<pair<int, int> > g[1000007];
long long node, k;
int vis[1000007];
int dfs(int u) {
  vis[u] = 1;
  int cc = 1;
  for (auto ch : g[u]) {
    if (ch.second == 0 && vis[ch.first] == 0) {
      cc += dfs(ch.first);
    }
  }
  return cc;
}
int main(void) {
  scanf("%lld%lld", &node, &k);
  for (int i = 1; i < node; i++) {
    int x, y, z;
    scanf("%d%d%d", &x, &y, &z);
    g[x].push_back({y, z});
    g[y].push_back({x, z});
  }
  long long ans = BigMod(node, k);
  for (int i = 1; i <= node; i++) {
    if (vis[i] == 0) {
      long long res = dfs(i);
      ans = (ans - BigMod(res, k) + 1000000007) % 1000000007;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
