Boboniu has a directed graph with n vertices and m edges.

The out-degree of each vertex is at most k.

Each edge has an integer weight between 1 and m. No two edges have equal weights.

Boboniu likes to walk on the graph with some specific rules, which is represented by a tuple (c_1,c_2,…,c_k). If he now stands on a vertex u with out-degree i, then he will go to the next vertex by the edge with the c_i-th (1≤ c_i≤ i) smallest weight among all edges outgoing from u.

Now Boboniu asks you to calculate the number of tuples (c_1,c_2,…,c_k) such that

  * 1≤ c_i≤ i for all i (1≤ i≤ k). 
  * Starting from any vertex u, it is possible to go back to u in finite time by walking on the graph under the described rules. 

Input

The first line contains three integers n, m and k (2≤ n≤ 2⋅ 10^5, 2≤ m≤ min(2⋅ 10^5,n(n-1) ), 1≤ k≤ 9).

Each of the next m lines contains three integers u, v and w (1≤ u,v≤ n,u≠ v,1≤ w≤ m), denoting an edge from u to v with weight w. It is guaranteed that there are no self-loops or multiple edges and each vertex has at least one edge starting from itself.

It is guaranteed that the out-degree of each vertex is at most k and no two edges have equal weight.

Output

Print one integer: the number of tuples.

Examples

Input


4 6 3
4 2 1
1 2 2
2 4 3
4 1 4
4 3 5
3 1 6


Output


2


Input


5 5 1
1 4 1
5 1 2
2 5 3
4 3 4
3 2 5


Output


1


Input


6 13 4
3 5 1
2 5 2
6 3 3
1 4 4
2 6 5
5 3 6
4 1 7
4 3 8
5 2 9
4 2 10
2 1 11
6 1 12
4 6 13


Output


1

Note

For the first example, there are two tuples: (1,1,3) and (1,2,3). The blue edges in the picture denote the c_i-th smallest edges for each vertex, which Boboniu chooses to go through.

<image>

For the third example, there's only one tuple: (1,2,2,2).

<image>

The out-degree of vertex u means the number of edges outgoing from u.
#include <bits/stdc++.h>
std::map<std::pair<int32_t, int32_t>, std::set<std::pair<int32_t, int32_t> > >
    invalid;
bool compatible(std::pair<int32_t, int32_t> p1,
                std::pair<int32_t, int32_t> p2) {
  return invalid[p1].find(p2) == invalid[p1].end();
}
int32_t answer = 0;
void enumerate(int32_t step, int32_t num_steps, int32_t* choices) {
  for (int32_t i = 0; i < step; i++) {
    choices[step] = i;
    if (step == num_steps) {
      bool ok = true;
      for (int32_t j = 1; j <= num_steps; j++) {
        for (int32_t k = j; k <= num_steps; k++) {
          if (!compatible({j, choices[j]}, {k, choices[k]})) {
            ok = false;
            break;
          }
        }
      }
      if (ok) answer++;
    } else
      enumerate(step + 1, num_steps, choices);
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  int32_t n, m, k;
  std::cin >> n >> m >> k;
  std::vector<std::pair<int32_t, int32_t> >* graph =
      new std::vector<std::pair<int32_t, int32_t> >[n];
  for (int32_t i = 0; i < m; i++) {
    int32_t src, dst, weight;
    std::cin >> src >> dst >> weight;
    src--;
    dst--;
    graph[src].emplace_back(weight, dst);
  }
  for (int32_t i = 0; i < n; i++) std::sort(graph[i].begin(), graph[i].end());
  std::set<std::pair<int32_t, int32_t> >* incoming_types =
      new std::set<std::pair<int32_t, int32_t> >[n];
  for (int32_t i = 0; i < n; i++) {
    for (int32_t j = 0; j < graph[i].size(); j++) {
      if (incoming_types[graph[i][j].second].find({graph[i].size(), j}) !=
          incoming_types[graph[i][j].second].end())
        invalid[{graph[i].size(), j}].emplace(graph[i].size(), j);
      incoming_types[graph[i][j].second].emplace(graph[i].size(), j);
    }
  }
  for (int32_t i = 1; i <= k; i++)
    for (int32_t j = 0; j < i; j++) {
      for (int32_t v = 0; v < n; v++)
        if (incoming_types[v].find({i, j}) != incoming_types[v].end())
          for (int32_t p = i; p <= k; p++)
            for (int32_t q = 0; q < p; q++) {
              if (incoming_types[v].find({p, q}) != incoming_types[v].end())
                if (i != p || j != q) invalid[{i, j}].emplace(p, q);
            }
    }
  int32_t* choices = new int32_t[k + 1];
  enumerate(1, k, choices);
  std::cout << answer << "\n";
  return 0;
}
