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 long long mod = 1e9 + 7;
const long long N = 200005;
inline long long read() {
  long long now = 0;
  char c = getchar();
  long long f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    now = (now << 1) + (now << 3) + (c ^ 48);
    c = getchar();
  }
  return f * now;
}
long long n, m, ans, head[N], tot, k;
bool vis[N];
struct node {
  long long v, next;
} edge[N << 1];
inline void add(long long x, long long y) {
  edge[++tot].next = head[x];
  edge[tot].v = y;
  head[x] = tot;
}
long long cnt = 0;
inline void dfs(long long now, long long fa) {
  vis[now] = 1;
  cnt++;
  for (long long i = head[now]; i; i = edge[i].next) {
    long long v = edge[i].v;
    if (v == fa || vis[v]) continue;
    dfs(v, now);
  }
}
signed main() {
  n = read();
  k = read();
  ans = pow(n, k);
  for (long long i = 1; i <= n - 1; i++) {
    long long x = read();
    long long y = read();
    long long z = read();
    if (z) continue;
    add(x, y);
    add(y, x);
  }
  for (long long i = 1; i <= n; i++) {
    if (vis[i]) continue;
    cnt = 0;
    dfs(i, 0);
    ans = (ans - pow(cnt, k) + mod);
    ans %= mod;
  }
  printf("%lld\n", ans);
  return 0;
}
