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>
using namespace std;
using ll = long long;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<pair<int, int>>> g(n);
  for (int i = 0; i < m; i++) {
    int v, u, w;
    cin >> v >> u >> w;
    v--, u--;
    g[v].emplace_back(u, w);
  }
  vector<vector<pair<int, int>>> into(n);
  for (int i = 0; i < n; i++) {
    if (!g[i].size()) return cout << "0\n", 0;
    sort(g[i].begin(), g[i].end(), [](pair<int, int> &a, pair<int, int> &b) {
      return a.second < b.second;
    });
    for (int j = 0; j < g[i].size(); j++)
      into[g[i][j].first].emplace_back(g[i].size() - 1, j);
  }
  vector<vector<char>> valid(k * k, vector<char>(k * k, 1));
  vector<vector<char>> can(k, vector<char>(k, 1));
  for (int v = 0; v < n; v++) {
    if (!into[v].size()) return cout << "0\n", 0;
    sort(into[v].begin(), into[v].end());
    vector<pair<int, int>> ninto;
    for (int i = 0; i < into[v].size();) {
      int j = i + 1;
      while (j < into[v].size() && into[v][i] == into[v][j]) j++;
      if (j - i > 1)
        can[into[v][i].first][into[v][i].second] = 0;
      else
        ninto.push_back(into[v][i]);
      i = j;
    }
    into[v] = ninto;
    if (!into[v].size()) return cout << "0\n", 0;
    for (int i = 0; i < into[v].size(); i++) {
      for (int j = i + 1; j < into[v].size(); j++) {
        pair<int, int> a = into[v][i], b = into[v][j];
        int ida = a.first * k + a.second;
        int idb = b.first * k + b.second;
        valid[ida][idb] = valid[idb][ida] = 0;
      }
    }
  }
  int ans = 0;
  vector<int> c;
  c.reserve(k);
  function<void()> dfs = [&]() {
    if (c.size() == k) {
      for (int i = 0; i < k; i++) {
        int idi = i * k + c[i];
        for (int j = i + 1; j < k; j++) {
          int idj = j * k + c[j];
          if (!valid[idi][idj]) return;
        }
      }
      ans++;
      return;
    }
    for (int now = 0; now <= c.size(); now++) {
      if (!can[c.size()][now]) continue;
      c.push_back(now);
      dfs();
      c.pop_back();
    }
  };
  dfs();
  cout << ans << '\n';
}
