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 M = 1000000007;
int n, m, k;
vector<pair<pair<int, int>, int>> edg;
vector<vector<pair<int, int>>> z;
vector<int> deg;
vector<vector<int>> gt;
int ans = 0;
vector<int> nabor;
set<pair<pair<int, int>, pair<int, int>>> ban;
set<pair<int, int>> permach;
void brute(int i) {
  if (i == k + 1) {
    ans++;
    return;
  }
  for (int kto = 0; kto < i; kto++) {
    if (permach.find({i, kto}) == permach.end()) {
      nabor[i] = kto;
      bool ok = true;
      for (int j = 0; j < i; j++) {
        if (ban.find({{j, nabor[j]}, {i, nabor[i]}}) != ban.end()) {
          ok = false;
          break;
        }
      }
      if (ok) {
        brute(i + 1);
      }
      nabor[i] = -1;
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n >> m >> k;
  z.resize(n);
  edg.resize(m);
  deg.resize(n);
  gt.resize(n);
  nabor.resize(k + 1);
  for (int i = 0; i < m; i++) {
    int a, b, w;
    cin >> a >> b >> w;
    a--;
    b--;
    edg.push_back({{a, b}, w});
    gt[b].push_back(i);
    deg[a]++;
    z[a].push_back({w, i});
  }
  for (int v = 0; v < n; v++) {
    sort(z[v].begin(), z[v].end());
  }
  vector<pair<int, int>> ex(m);
  for (int v = 0; v < n; v++) {
    for (int i = 0; i < z[v].size(); i++) {
      ex[z[v][i].second] = {deg[v], i};
    }
  }
  for (int v = 0; v < n; v++) {
    set<pair<int, int>> zlo;
    for (auto i : gt[v]) {
      if (zlo.find(ex[i]) != zlo.end()) permach.insert(ex[i]);
      zlo.insert(ex[i]);
    }
    vector<pair<int, int>> vg;
    for (auto p : zlo) vg.push_back(p);
    for (int i = 0; i < vg.size(); i++) {
      for (int j = i + 1; j < vg.size(); j++) {
        ban.insert({vg[i], vg[j]});
        ban.insert({vg[j], vg[i]});
      }
    }
  }
  brute(1);
  cout << ans << '\n';
}
