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;
template <class T>
ostream& operator<<(ostream& out, vector<T>& v) {
  out << "[";
  for (auto k : v) out << k << " ";
  out << "]"
      << "\n";
  return out;
}
template <class T>
ostream& operator<<(ostream& out, set<T> s) {
  out << "{";
  for (auto k : s) out << k << " ";
  out << "}"
      << "\n";
  return out;
}
template <class T, class U>
ostream& operator<<(ostream& out, pair<T, U> p) {
  out << "[ " << p.first << " , " << p.second << " ] ";
  return out;
}
template <class T, class U>
istream& operator>>(istream& in, pair<T, U>& p) {
  in >> p.first >> p.second;
  return in;
}
int cant[10], ind[10], N, M, K, totf = 0;
vector<int> mat[10][10];
vector<pair<int, int> > g[300000];
int in[200000];
bitset<100> acum;
bitset<100> mal[100];
void Recu(int i) {
  if (i == K) {
    totf++;
    return;
  }
  int j = ind[i];
  for (int k = int(1); k < int(j + 1); k++) {
    bool good = true;
    int id = j * 10 + k;
    acum[id] = true;
    if ((acum & mal[id]).count()) good = false;
    if (good) Recu(i + 1);
    acum[id] = false;
  }
}
bool Comp(int i, int j) { return cant[i] > cant[j]; }
bitset<100> comps[300000];
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  cin >> N >> M >> K;
  for (int i = 0; i < int(M); i++) {
    int a, b, c;
    cin >> a >> b >> c;
    g[a].push_back({c, b});
  }
  for (int i = int(1); i < int(N + 1); i++) {
    sort(g[i].begin(), g[i].end());
    for (int j = 0; j < int(g[i].size()); j++) {
      int nod = g[i][j].second;
      int id = 10 * g[i].size() + j + 1;
      for (int x = 0; x < int(100); x++)
        if (comps[nod][x]) mal[x][id] = mal[id][x] = true;
      comps[nod][id] = true;
    }
    cant[g[i].size()]++;
  }
  for (int i = 0; i < int(9); i++) ind[i] = i + 1;
  sort(ind, ind + K, Comp);
  Recu(0);
  cout << totf << "\n";
  return 0;
}
