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;
const long long N = 2e5 + 69;
signed main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  long long n, m, k;
  cin >> n >> m >> k;
  vector<vector<pair<long long, long long>>> adj(n);
  for (long long i = 0; i < m; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    u--;
    v--;
    adj[u].emplace_back(w, v);
  }
  vector<vector<bitset<N>>> disc(11);
  for (long long i = 1; i <= 10; i++) {
    disc[i] = vector<bitset<N>>(i);
  }
  vector<vector<long long>> ct(11, vector<long long>(11));
  for (long long u = 0; u < n; u++) {
    if (0) cerr << "u is " << u + 1 << "\n";
    sort(adj[u].begin(), adj[u].end());
    long long deg = adj[u].size();
    for (long long i = 0; i < (long long)adj[u].size(); i++) {
      long long v = adj[u][i].second;
      if (0)
        cerr << "insert " << v + 1 << " into "
             << "disc[" << deg << "][" << i + 1 << "]\n";
      if (disc[deg][i][v]) ct[deg][i] = -1;
      if (ct[deg][i] != -1) ct[deg][i]++;
      disc[deg][i][v] = true;
    }
  }
  bool ok[11][11][11][11];
  for (long long i1 = 1; i1 <= 10; i1++) {
    for (long long j1 = 0; j1 < i1; j1++) {
      for (long long i2 = 1; i2 <= 10; i2++) {
        for (long long j2 = 0; j2 < i2; j2++) {
          ok[i1][j1][i2][j2] = (ct[i1][j1] != -1) && (ct[i2][j2] != -1) &&
                               (disc[i1][j1] & disc[i2][j2]).none();
        }
      }
    }
  }
  vector<long long> p;
  long long ans = 0;
  function<void()> go = [&]() {
    if ((long long)p.size() == k) {
      bool add = true;
      long long tot = 0;
      for (long long i = 0; i < k; i++) {
        tot += ct[i + 1][p[i]];
        for (long long j = i + 1; j < k; j++) {
          if (!ok[i + 1][p[i]][j + 1][p[j]]) add = false;
        }
      }
      add &= tot == n;
      if (add) {
        if (0) cerr << "found sol\n";
        for (long long i = 0; i < k; i++) {
          if (0) cerr << p[i] + 1 << " \n"[i == k - 1];
        }
      }
      ans += add;
    } else {
      long long rn = p.size();
      for (long long i = 0; i <= rn; i++) {
        p.push_back(i);
        go();
        p.pop_back();
      }
    }
  };
  go();
  cout << ans << "\n";
}
