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 = 2 * 1000 * 100 + 10, C = 15;
int n, m, K, ans, deg[N], c[C];
vector<pair<int, int>> adj[N], adj1[N];
vector<int> vec[C], pos[C];
set<pair<int, int>> bad[C][C];
bool vis[N];
void bt(int i) {
  if (i > K) {
    ans++;
    return;
  }
  for (auto d : pos[i]) {
    bool go = true;
    for (auto it : bad[i][d])
      if (c[it.first] == it.second) go = false;
    if (go) {
      c[i] = d;
      bt(i + 1);
    }
  }
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m >> K;
  for (int i = 0, v, u, w; i < m; i++) {
    cin >> v >> u >> w;
    adj[v].push_back({w, u});
    deg[v]++;
  }
  for (int i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end());
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < deg[i]; j++) adj[i][j].first = j + 1;
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < deg[i]; j++)
      adj1[adj[i][j].second].push_back({deg[i], adj[i][j].first});
  for (int i = 1; i <= n; i++) vec[deg[i]].push_back(i);
  for (int d = 1; d <= K; d++)
    for (int x = 1; x <= d; x++) {
      set<int> s;
      for (auto v : vec[d]) s.insert(adj[v][x - 1].second);
      if ((int)(s.size()) == (int)(vec[d].size())) {
        pos[d].push_back(x);
        for (auto v : s)
          for (auto u : adj1[v])
            if (u.first < d) bad[d][x].insert({u.first, u.second});
      }
    }
  bt(1);
  cout << ans << '\n';
  return 0;
}
