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;
bool cmp(pair<int, int> l, pair<int, int> r) { return l.second < r.second; }
class Hash {
 public:
  long long a[4];
  static long long mod[4];
  Hash() {
    a[0] = a[2] = 0;
    a[1] = a[3] = 1;
  }
  void hash(Hash &b) {
    a[0] = (a[0] + b.a[0]) % mod[0];
    a[1] = (a[1] * b.a[1]) % mod[1];
    a[2] = (a[2] + b.a[2]) % mod[2];
    a[3] = (a[3] * b.a[3]) % mod[3];
  }
  void hash(int b) {
    a[0] = (a[0] + b) % mod[0];
    a[1] = (a[1] * b) % mod[1];
    a[2] = (a[2] + b) % mod[2];
    a[3] = (a[3] * b) % mod[3];
  }
  bool is_valid(Hash &b) {
    for (int i = 0; i < 4; ++i)
      if (a[i] != b.a[i]) return false;
    return true;
  }
};
long long Hash::mod[4];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m, k;
  vector<bool> vi;
  vector<vector<pair<int, int>>> edge;
  vector<vector<int>> kv;
  cin >> n >> m >> k;
  edge.resize(n + 1);
  vi.resize(n + 1);
  for (int i = 1; i <= n; ++i) vi[i] = false;
  int u, v, w;
  for (int i = 0; i < m; ++i) {
    cin >> u >> v >> w;
    edge[u].push_back(make_pair(v, w));
  }
  for (int i = 1; i <= n; ++i) sort(edge[i].begin(), edge[i].end(), cmp);
  kv.resize(k + 1);
  for (int i = 1; i <= n; ++i) kv[edge[i].size()].push_back(i);
  Hash::mod[0] = 1000000007ll;
  Hash::mod[1] = 999999943ll;
  Hash::mod[2] = 2000000029ll;
  Hash::mod[3] = 1999999922ll;
  vector<vector<Hash>> h;
  h.resize(k + 1);
  for (int i = 1; i <= k; ++i) {
    h[i].resize(i + 1);
    for (int j = 1; j <= i; ++j) {
      for (int t : kv[i]) {
        h[i][j].hash(edge[t][j - 1].first);
      }
    }
  }
  Hash gt;
  for (int i = 1; i <= n; ++i) gt.hash(i);
  vector<int> per;
  per.resize(k + 1);
  for (int i = 1; i <= k; ++i) per[i] = 1;
  int ans = 0;
  while (true) {
    Hash hh;
    for (int i = 1; i <= k; ++i) hh.hash(h[i][per[i]]);
    if (gt.is_valid(hh)) ++ans;
    int t = 1;
    while (t <= k && per[t] == t) ++t;
    if (t > k)
      break;
    else {
      ++per[t];
      for (int i = 1; i < t; ++i) per[i] = 1;
    }
  }
  cout << ans << endl;
  return 0;
}
