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;
int n, m, k;
const int N = 2e5 + 5;
long long pr[N], P[N];
vector<pair<int, int> > adj[N], e[N], F[N];
int d[N][10];
vector<int> deg[10][10];
int A[10][10][10][10];
int cur, ans;
vector<pair<int, int> > chosen;
vector<vector<int> > t;
void gen() {
  cur++;
  if (cur > k) {
    ans++;
    return;
  }
  for (int i = 1; i <= cur; i++) {
    if (A[cur][i][cur][i]) continue;
    bool flag = 1;
    for (auto& z : chosen) {
      if (A[cur][i][z.first][z.second]) {
        flag = 0;
        break;
      }
    }
    if (flag) {
      chosen.push_back({cur, i});
      gen();
      cur--;
      chosen.pop_back();
    }
  }
}
int W[] = {1, 1, 2, 1};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0, a, b, w; i < m; i++) {
    cin >> a >> b >> w;
    adj[a].push_back({w, b});
    e[b].push_back({w, a});
  }
  for (int i = 1; i <= n; i++) {
    sort(e[i].begin(), e[i].end(), [&](pair<int, int> a, pair<int, int> b) {
      return (adj[a.second].size() == adj[b.second].size()
                  ? a < b
                  : adj[a.second].size() < adj[b.second].size());
    });
    sort(adj[i].begin(), adj[i].end());
  }
  for (int i = 1; i <= n; i++) {
    for (int c = 1; c <= adj[i].size(); c++) {
      F[adj[i][c - 1].second].push_back({adj[i].size(), c});
    }
  }
  for (int i = 1; i <= n; i++) {
    sort(F[i].begin(), F[i].end());
    for (int j = 1; j < F[i].size(); j++)
      if (F[i][j] == F[i][j - 1])
        A[F[i][j].first][F[i][j].second][F[i][j].first][F[i][j].second] = 1;
    F[i].resize(unique((F[i]).begin(), (F[i]).end()) - F[i].begin());
    for (int j = 0; j < F[i].size(); j++)
      for (int c = 0; c < F[i].size(); c++) {
        if (j == c) continue;
        auto a = F[i][j];
        auto b = F[i][c];
        A[a.first][a.second][b.first][b.second] = 1;
      }
  }
  gen();
  cout << ans;
}
