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 c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
mt19937 myrand(time(0));
long long n, m, k, ans;
vector<pair<long long, long long> > out[200002], in[200002];
bool bad[50][50];
long long id[50][50];
long long getid(long long i, long long j) { return (i * (i + 1) / 2) + j; }
void rec(long long pos, long long mask) {
  if (pos > k) {
    ans++;
    return;
  }
  for (int i = 0; i < pos; i++) {
    if (bad[pos][i] || mask & id[pos][i]) continue;
    rec(pos + 1, mask | (1ll << getid(pos, i)));
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(), cout.tie();
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    out[u].push_back({w, v});
  }
  for (int i = 1; i <= n; i++) {
    sort(out[i].begin(), out[i].end());
    if (out[i].empty()) {
      cout << 0 << endl;
      return 0;
    }
    for (int j = 0; j < (int)out[i].size(); j++) {
      in[out[i][j].second].push_back({out[i].size(), j});
    }
  }
  for (int i = 1; i <= n; i++) {
    sort(in[i].begin(), in[i].end());
    vector<pair<long long, long long> > tmp;
    for (int j = 0; j < (int)in[i].size(); j++) {
      if (!tmp.empty() && tmp.back() == in[i][j]) {
        bad[in[i][j].first][in[i][j].second] = 1;
        tmp.pop_back();
      } else
        tmp.push_back(in[i][j]);
    }
    for (int j = 0; j < (int)tmp.size(); j++) {
      for (int l = 0; l < (int)tmp.size(); l++) {
        id[tmp[j].first][tmp[j].second] |=
            (1ll << getid(tmp[l].first, tmp[l].second));
      }
    }
  }
  rec(1, 0);
  cout << ans << endl;
}
