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 int N = 2e5 + 5, A = 15;
int ans, n, m, k, p[A];
vector<pair<int, int> > G[N];
int c[N][A][A];
bool b[A][A], b2[A][A][A][A];
void DFS(int x) {
  if (x == k + 1) {
    for (int i = 1; i <= k; i++)
      if (b[i][p[i]]) return;
    for (int i = 1; i <= k; i++)
      for (int j = i + 1; j <= k; j++)
        if (b2[i][p[i]][j][p[j]]) return;
    ans++;
    return;
  }
  for (int i = 1; i <= x; i++) {
    p[x] = i;
    DFS(x + 1);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    G[u].push_back(make_pair(w, v));
  }
  for (int u = 1; u <= n; u++) {
    if (!G[u].size()) {
      cout << 0 << endl;
      return 0;
    }
    sort(G[u].begin(), G[u].end());
    int cnt = 0;
    for (auto i : G[u]) c[i.second][G[u].size()][++cnt]++;
  }
  for (int u = 1; u <= n; u++) {
    vector<pair<int, int> > now;
    for (int i = 1; i <= k; i++)
      for (int j = 1; j <= i; j++) {
        if (c[u][i][j] >= 2) {
          b[i][j] = true;
          continue;
        }
        if (c[u][i][j] == 1) now.push_back(make_pair(i, j));
      }
    for (unsigned int i = 0; i < now.size(); i++)
      for (unsigned int j = i + 1; j < now.size(); j++)
        b2[now[i].first][now[i].second][now[j].first][now[j].second] = true;
  }
  DFS(1);
  cout << ans << endl;
  return 0;
}
