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, class V>
ostream &operator<<(ostream &s, pair<T, V> a) {
  s << a.first << ' ' << a.second;
  return s;
}
int invalid[10][10];
int invalid2[10][10][10][10];
vector<pair<long long int, long long int> > adj[200005];
set<pair<long long int, long long int> > inp[200005];
long long int ans = 0;
int n, m, k;
void solve(vector<long long int> &temp, int depth) {
  if (depth == k) {
    bool ok = true;
    for (int i = 0; ok && i < k; i++) {
      if (invalid[i + 1][temp[i]]) ok = false;
      for (int j = i + 1; j < k && ok; j++) {
        if (invalid2[i + 1][temp[i]][j + 1][temp[j]]) ok = false;
      }
    }
    if (ok) ans++;
    return;
  }
  for (int i = 0; i <= depth; i++) {
    temp.push_back(i);
    solve(temp, depth + 1);
    temp.pop_back();
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int T = 1;
  for (int qq = 1; qq <= T; qq++) {
    cin >> n >> m >> k;
    for (int i = 0; i < m; i++) {
      int u, v, w;
      cin >> u >> v >> w;
      adj[u].push_back(pair<long long int, long long int>(w, v));
    }
    for (int i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end());
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < adj[i].size(); j++) {
        if (!inp[adj[i][j].second]
                 .insert(pair<long long int, long long int>(adj[i].size(), j))
                 .second)
          invalid[adj[i].size()][j] = 1;
      }
    }
    for (int i = 1; i <= n; i++) {
      for (auto x : inp[i]) {
        for (auto y : inp[i]) {
          invalid2[x.first][x.second][y.first][y.second] = 1;
          invalid2[y.first][y.second][x.first][x.second] = 1;
        }
      }
    }
    vector<long long int> t;
    solve(t, 0);
    cout << ans << endl;
  }
}
