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>
void _R(T &x) {
  cin >> x;
}
void _R(signed &x) { scanf("%d", &x); }
void _R(int64_t &x) { scanf("%lld", &x); }
void _R(double &x) { scanf("%lf", &x); }
void _R(char &x) { scanf(" %c", &x); }
void _R(char *x) { scanf("%s", x); }
void R() {}
template <class T, class... U>
void R(T &head, U &...tail) {
  _R(head);
  R(tail...);
}
template <class T>
void _W(const T &x) {
  cout << x;
}
void _W(const signed &x) { printf("%d", x); }
void _W(const int64_t &x) { printf("%lld", x); }
void _W(const double &x) { printf("%.16f", x); }
void _W(const char &x) { putchar(x); }
void _W(const char *x) { printf("%s", x); }
template <class T, class U>
void _W(const pair<T, U> &x) {
  _W(x.first);
  putchar(' ');
  _W(x.second);
}
template <class T>
void _W(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _W(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void W() {}
template <class T, class... U>
void W(const T &head, const U &...tail) {
  _W(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  W(tail...);
}
const int N = 2e5 + 10, D = 9;
int n, m, k;
vector<pair<int, int> > adj[N];
int cnt[N][D][D];
bool no[D][D][D][D], noo[D][D];
int p[D];
int ans;
void dfs(int dep) {
  if (dep == k) {
    ans++;
    return;
  }
  for (int i = (0); i < (dep + 1); ++i)
    if (!noo[dep][i]) {
      p[dep] = i;
      bool flag = false;
      for (int j = (0); j < (dep); ++j) {
        flag |= no[dep][p[dep]][j][p[j]];
      }
      if (!flag) dfs(dep + 1);
    }
}
signed main() {
  R(n, m, k);
  for (int i = (0); i < (m); ++i) {
    int u, v, w;
    R(u, v, w);
    adj[u].push_back(make_pair(w, v));
  }
  for (int i = (1); i < (n + 1); ++i) {
    sort((adj[i]).begin(), (adj[i]).end());
    for (int k = (0); k < (adj[i].size()); ++k) {
      int j = adj[i][k].second;
      cnt[j][(int)adj[i].size() - 1][k]++;
    }
  }
  for (int i = (1); i < (n + 1); ++i) {
    for (int j = (0); j < (D); ++j) {
      for (int b1 = (0); b1 < (j + 1); ++b1)
        if (cnt[i][j][b1]) {
          if (cnt[i][j][b1] > 1) {
            noo[j][b1] = 1;
            continue;
          }
          for (int k = (0); k < (j); ++k) {
            for (int b2 = (0); b2 < (k + 1); ++b2)
              if (cnt[i][k][b2] == 1) {
                no[j][b1][k][b2] = 1;
              }
          }
        }
    }
  }
  dfs(0);
  W(ans);
}
