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 <typename T, typename S>
ostream &operator<<(ostream &os, const pair<T, S> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename C, typename T = decay<decltype(*begin(declval<C>()))>,
          typename enable_if<!is_same<C, string>::value>::type * = nullptr>
ostream &operator<<(ostream &os, const C &c) {
  bool f = true;
  os << "[";
  for (const auto &x : c) {
    if (!f) os << ", ";
    f = false;
    os << x;
  }
  return os << "]";
}
template <typename T>
void debug(string s, T x) {
  cerr << s << " = " << x << "\n";
}
template <typename T, typename... Args>
void debug(string s, T x, Args... args) {
  cerr << s.substr(0, s.find(',')) << " = " << x << " | ";
  debug(s.substr(s.find(',') + 2), args...);
}
int k, has[10][10];
bool ban[10][10][10][10], ban2[10][10];
vector<int> cur;
vector<pair<int, int>> adj[200005], pairs[200005];
int solve(int i) {
  if (i == k + 1) return 1;
  int ret = 0;
  for (int j = 0; j < i; j++)
    if (!ban2[i][j]) {
      bool ok = true;
      for (int l = 1; l < i; l++) ok &= !ban[i][j][l][cur[l - 1]];
      if (ok) {
        cur.push_back(j);
        ret += solve(i + 1);
        cur.pop_back();
      }
    }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    u--, v--;
    adj[u].emplace_back(v, w);
  }
  for (int i = 0; i < n; i++) {
    sort(adj[i].begin(), adj[i].end(),
         [](const pair<int, int> &a, const pair<int, int> &b) {
           return a.second < b.second;
         });
    for (int j = 0; j < adj[i].size(); j++)
      pairs[adj[i][j].first].emplace_back(adj[i].size(), j);
  }
  for (int i = 0; i < n; i++) {
    for (auto &p : pairs[i]) {
      if (has[p.first][p.second] == i + 1)
        ban2[p.first][p.second] = true;
      else
        has[p.first][p.second] = i + 1;
    }
    for (int a = 1; a <= k; a++)
      for (int b = 0; b < a; b++)
        if (has[a][b] == i + 1)
          for (int c = 1; c <= k; c++)
            for (int d = 0; d < c; d++)
              if (has[c][d] == i + 1) ban[a][b][c][d] = ban[c][d][a][b] = true;
  }
  cout << solve(1) << "\n";
  return 0;
}
