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 INF = 2147483647;
const long long INFL = 9223372036854775807LL;
const double EPSILON = 0.00000001;
const long long MOD = 1000000007;
vector<long long> adj[100000 + 5];
bool visited[100000 + 5];
long long mod_pow(long long b, long long e, long long mod) {
  long long r = 1, q = b;
  while (e) {
    if (e & 1) r = (r * q) % mod;
    q = (q * q) % mod;
    e >>= 1;
  }
  return r;
}
long long search(long long start) {
  long long cnt = 0;
  queue<long long> q;
  q.push(start);
  visited[start] = true;
  while (not q.empty()) {
    long long top = q.front();
    q.pop();
    cnt++;
    for (long long x : adj[top]) {
      if (not visited[x]) {
        visited[x] = true;
        q.push(x);
      }
    }
  }
  return cnt;
}
int32_t main() {
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v, x;
    scanf("%I64d", &u);
    ;
    scanf("%I64d", &v);
    ;
    scanf("%I64d", &x);
    ;
    u--;
    v--;
    if (x == 0) {
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
  }
  long long tot = mod_pow(n, k, MOD);
  long long bad = 0;
  memset(visited, false, sizeof(visited));
  ;
  for (long long i = 0; i < n; i++) {
    if (not visited[i]) {
      long long cnt = search(i);
      bad += mod_pow(cnt, k, MOD);
      bad %= MOD;
    }
  }
  long long good = tot - bad;
  cout << (((good) % (MOD) + MOD) % MOD) << endl;
  return 0;
}
