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 int MOD = 1000000007;
vector<vector<long long int> > graph;
vector<int> visited;
vector<int> level;
vector<int> colors;
long long int multmod(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  long long int ans = 0;
  while (b > 0) {
    if (b & 1 == 1) {
      ans = (ans + a) % m;
    }
    b = b >> 1;
    a = (a * 2) % m;
  }
  return ans % m;
}
long long int powmod(long long int x, long long int y, long long int m) {
  if (x == 0) {
    return 0;
  }
  long long int ans = 1;
  x = x % m;
  while (y > 0) {
    if (y & 1 == 1) {
      ans = multmod(ans, x, m) % m;
    }
    x = multmod(x, x, m) % m;
    y = y >> 1;
  }
  return ans % m;
}
long long int nodesnum;
void dfs_stack(long long int node) {
  long long int top;
  stack<long long int> nodes;
  nodes.push(node);
  visited[node] = 1;
  while (!nodes.empty()) {
    top = nodes.top();
    nodes.pop();
    nodesnum++;
    for (long long int i : graph[top]) {
      if (!visited[i]) {
        nodes.push(i);
        visited[i] = 1;
      }
    }
  }
}
void bfs(long long int node) {
  long long int top;
  queue<long long int> nodes;
  nodes.push(node);
  visited[node] = 1;
  level[node] = 0;
  while (!nodes.empty()) {
    top = nodes.front();
    nodes.pop();
    cout << top << endl;
    for (long long int i : graph[top]) {
      if (!visited[i]) {
        nodes.push(i);
        visited[i] = 1;
        level[i] = level[top] + 1;
      }
    }
  }
}
int main() {
  long long int n, k;
  cin >> n >> k;
  graph.resize(n);
  visited.resize(n);
  long long int u, v, x;
  for (long long int i = 0; i < n - 1; i++) {
    cin >> u >> v >> x;
    if (x == 0) {
      graph[u - 1].push_back(v - 1);
      graph[v - 1].push_back(u - 1);
    }
  }
  nodesnum = 0;
  long long int sub = 0;
  for (long long int i = 0; i < n; i++) {
    if (!visited[i]) {
      nodesnum = 0;
      dfs_stack(i);
      sub = (sub + powmod(nodesnum, k, 1000000007)) % 1000000007;
    }
  }
  long long int ans = powmod(n, k, 1000000007) % 1000000007;
  ans = (ans + MOD - sub) % 1000000007;
  cout << ans % 1000000007;
  return 0;
}
