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;
using ll = long long;
using vi = vector<int>;
using vl = vector<long long>;
using pii = pair<int, int>;
using vpii = vector<pair<int, int>>;
template <class T>
inline bool ckmin(T& a, T b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
inline bool ckmax(T& a, T b) {
  return b > a ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const char nl = '\n';
const int mxN = 2e5 + 10;
const int MOD = 1e9 + 7;
const long long infLL = 1e18;
int n, m, k;
vpii adj[mxN];
map<pii, int> mp[mxN];
int hsh(pii q) { return q.first * 10 + q.second; }
bool bad[500][500];
int ans = 0;
vpii stk;
void dfs(int x) {
  if (x == k + 1) return (void)ans++;
  for (int i = (0); i < (x); i++) {
    pii p = {x, i};
    stk.push_back(p);
    bool meh = 0;
    for (auto& v : stk)
      if (bad[hsh(p)][hsh(v)]) meh = 1;
    if (!meh) {
      dfs(x + 1);
    }
    stk.pop_back();
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = (0); i < (m); i++) {
    int u, v, w;
    cin >> u >> v >> w;
    u--;
    v--;
    adj[u].push_back({w, v});
  }
  for (int i = (0); i < (n); i++) {
    sort(adj[i].begin(), adj[i].end());
    for (int j = (0); j < ((int)adj[i].size()); j++) {
      mp[adj[i][j].second][{(int)adj[i].size(), j}]++;
    }
  }
  for (int i = (0); i < (n); i++) {
    for (auto& x : mp[i]) {
      for (auto& y : mp[i]) {
        if (x.first == y.first) {
          if (x.second > 1)
            bad[hsh(x.first)][hsh(y.first)] = 1,
            bad[hsh(y.first)][hsh(x.first)] = 1;
        } else {
          bad[hsh(x.first)][hsh(y.first)] = 1;
          bad[hsh(y.first)][hsh(x.first)] = 1;
        }
      }
    }
  }
  dfs(1);
  cout << ans << nl;
  return 0;
}
