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 MaxN = 2e5 + 10;
int n, m, k;
vector<pair<int, int>> Adj[MaxN];
map<set<int>, int> mp1, mp2;
vector<int> s[10];
vector<vector<int>> d[10];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  int u, v, w;
  for (int i = 1; i <= m; i++) {
    cin >> u >> v >> w;
    Adj[u].push_back(make_pair(v, w));
  }
  for (int i = 1; i <= n; i++) {
    sort(Adj[i].begin(), Adj[i].end(), [&](pair<int, int> x, pair<int, int> y) {
      return x.second < y.second;
    });
  }
  for (int i = 1; i <= n; i++) {
    s[Adj[i].size()].push_back(i);
  }
  for (int i = 1; i <= k; i++) {
    if (s[i].empty()) {
      continue;
    }
    for (int j = 0; j < i; j++) {
      vector<int> tmp;
      for (auto &v : s[i]) {
        tmp.push_back(Adj[v][j].first);
      }
      sort(tmp.begin(), tmp.end());
      tmp.resize(unique(tmp.begin(), tmp.end()) - tmp.begin());
      if (tmp.size() == s[i].size()) {
        d[i].push_back(tmp);
      }
    }
  }
  mp1[set<int>()] = 1;
  for (int i = 1; i <= k; i++) {
    if (s[i].empty()) {
      for (auto &p : mp1) {
        p.second *= i;
      }
      continue;
    }
    for (auto &vec : d[i]) {
      for (auto &p : mp1) {
        set<int> tmp = p.first;
        bool ok = 1;
        for (auto &v : vec) {
          if (tmp.count(v)) {
            ok = 0;
            break;
          }
          tmp.insert(v);
        }
        if (ok) {
          mp2[tmp] += p.second;
        }
      }
    }
    mp1 = mp2;
    mp2.clear();
  }
  cout << (mp1.empty() ? 0 : (*mp1.begin()).second);
  return 0;
}
