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;
int ans = 0;
vector<set<pair<int, int>>> fm;
int n, m, k;
set<pair<int, int>> self_con;
map<pair<int, int>, set<pair<int, int>>> con;
vector<vector<int>> out;
vector<vector<pair<int, int>>> ns;
vector<pair<int, int>> cur;
void dfs(int d) {
  if (d == k + 1) {
    ans += 1;
    return;
  }
  for (int i = 0; i < d; i++) {
    if (self_con.find(make_pair(i, d - 1)) != self_con.end()) continue;
    int rs = 1;
    pair<int, int> mkp = make_pair(i, d - 1);
    for (auto it = cur.begin(); it != cur.end(); it++) {
      if (con[(*it)].find(mkp) != con[(*it)].end()) {
        rs = 0;
        break;
      }
    }
    if (rs == 1) {
      cur.push_back(mkp);
      dfs(d + 1);
      cur.pop_back();
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  cin >> n >> m >> k;
  fm.resize(n);
  ns.resize(n);
  for (int i = 0; i < m; i++) {
    int f, t, w;
    cin >> f >> t >> w;
    f--;
    t--;
    ns[f].push_back(make_pair(w, t));
  }
  for (int i = 0; i < n; i++) {
    sort(ns[i].begin(), ns[i].end());
    int ll = ns[i].size();
    for (int j = 0; j < ll; j++) {
      int t = ns[i][j].second;
      pair<int, int> mkp = make_pair(j, ll - 1);
      if (fm[t].find(mkp) != fm[t].end())
        self_con.insert(mkp);
      else
        fm[t].insert(mkp);
    }
  }
  for (int i = 0; i < n; i++) {
    for (auto j = fm[i].begin(); j != fm[i].end(); j++) {
      auto k = j;
      k++;
      for (; k != fm[i].end(); k++) {
        con[*j].insert(*k);
        con[*k].insert(*j);
      }
    }
  }
  dfs(1);
  cout << ans << endl;
  return 0;
}
