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 uint = unsigned int;
using cat = long long;
using dbl = long double;
constexpr dbl pi = 3.14159265358979323846;
using namespace std;
template <typename T>
T abs(T x) {
  return (x < 0) ? (-x) : x;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(10);
  int N, M, K;
  cin >> N >> M >> K;
  vector<vector<pair<int, int> > > E_out(N);
  for (int i = 0; i < M; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    E_out[u - 1].push_back({w, v - 1});
  }
  vector<vector<pair<int, int> > > E_in(N);
  for (int i = 0; i < N; i++) {
    sort(begin(E_out[i]), end(E_out[i]));
    for (int j = 0; j < (int)E_out[i].size(); j++)
      E_in[E_out[i][j].second].push_back({E_out[i].size(), j});
  }
  char bl[10][10];
  memset(bl, 0, sizeof(bl));
  char bl_p[10][10][10][10];
  memset(bl_p, 0, sizeof(bl_p));
  int cnt[10][10];
  memset(cnt, 0, sizeof(cnt));
  for (int i = 0; i < N; i++) {
    char occ[10][10];
    memset(occ, 0, sizeof(occ));
    for (auto p : E_in[i]) occ[p.first][p.second]++;
    for (int j = 1; j <= K; j++)
      for (int k = 0; k < j; k++) {
        if (occ[j][k] > 1) bl[j][k] = 1;
        if (occ[j][k] == 1) {
          cnt[j][k]++;
          for (int l = j + 1; l <= K; l++)
            for (int m = 0; m < l; m++)
              if (occ[l][m] == 1) bl_p[j][k][l][m] = 1;
        }
      }
  }
  int P_cnt = 1;
  for (int i = 1; i <= K; i++) P_cnt *= i;
  int ans = 0;
  for (int i = 0; i < P_cnt; i++) {
    int st[10];
    for (int j = 0, a = i; j < K; j++) {
      st[j] = a % (j + 1);
      a /= j + 1;
    }
    int sum_cnt = 0;
    bool ok = true;
    char is_bl[10][10];
    memset(is_bl, 0, sizeof(is_bl));
    for (int j = 0; j < K; j++) {
      if (bl[j + 1][st[j]] || is_bl[j + 1][st[j]]) {
        ok = false;
        break;
      }
      for (int k = j + 1; k < K; k++)
        for (int l = 0; l <= k; l++)
          is_bl[k + 1][l] |= bl_p[j + 1][st[j]][k + 1][l];
      sum_cnt += cnt[j + 1][st[j]];
    }
    if (ok && sum_cnt == N) ans++;
  }
  cout << ans << "\n";
}
