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;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const int maxn = 2e5 + 10;
const long long mod = 1e9 + 7;
int head[maxn], cnt = 1, cc = 0;
int vis[maxn];
struct tree {
  int u, v, x, next;
} es[maxn * 2];
void add(int u, int v, int x) {
  es[cnt].u = u;
  es[cnt].v = v;
  es[cnt].x = x;
  es[cnt].next = head[u];
  head[u] = cnt++;
}
int dfs(int s, int past) {
  vis[s] = 1;
  cc++;
  for (int i = head[s]; i != 0; i = es[i].next) {
    int v = es[i].v;
    if (v == past) continue;
    if (es[i].x == 0) {
      dfs(v, s);
    }
  }
}
long long fpow(long long a, long long n, long long p) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = ans * a % p;
    a = a * a % p;
    n >>= 1;
  }
  return ans;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int u, v, x;
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d%d%d", &u, &v, &x);
    add(u, v, x);
    add(v, u, x);
  }
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    if (vis[i]) continue;
    cc = 0;
    dfs(i, 0);
    sum += fpow(cc, k, mod);
  }
  cout << fpow(n, k, mod) - sum << endl;
}
