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 = (int)2e5 + 100;
const int K = 9;
const int Z = K * K;
bitset<Z> nx[K][K];
vector<pair<int, int>> I[N];
vector<pair<int, int>> T[N];
int force[K];
int has[K][K];
int k;
vector<vector<pair<int, int>>> conds;
int answ = 0;
void brute(vector<int> cc) {
  if (cc.size() == k) {
    for (int i = 0; i < k; i++) {
      for (int a = 0; a < k; a++) {
        if (nx[i][cc[i]][a * k + cc[a]]) return;
      }
    }
    int cnt = 0;
    for (int i = 0; i < k; i++) cnt += has[i][cc[i]];
    if (cnt != conds.size()) return;
    answ++;
    return;
  }
  int f = cc.size();
  for (int i = 0; i <= f; i++) {
    if (force[f] != -1 && force[f] != i) continue;
    cc.push_back(i);
    brute(cc);
    cc.pop_back();
  }
}
int num[K][K];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, m;
  cin >> n >> m >> k;
  int u, v, w;
  for (int j = 0; j < k; j++) {
    force[j] = -1;
  }
  for (int i = 0; i < m; i++) {
    cin >> u >> v >> w;
    T[u].push_back(make_pair(w, v));
  }
  for (int i = 1; i <= n; i++) {
    sort(T[i].begin(), T[i].end());
    for (int j = 0; j < T[i].size(); j++) {
      I[T[i][j].second].push_back(make_pair((int)T[i].size() - 1, j));
    }
  }
  for (int i = 1; i <= n; i++) {
    if (I[i].size() == 0) continue;
    if (I[i].size() == 1) {
      if (force[I[i][0].first] != -1) {
        if (force[I[i][0].first] != I[i][0].second) {
          cout << "0\n";
          return 0;
        }
      }
      force[I[i][0].first] = I[i][0].second;
    } else {
      bitset<Z> vse;
      for (auto p : I[i]) {
        num[p.first][p.second]++;
        vse[p.first * k + p.second] = 1;
      }
      for (auto p : I[i]) {
        if (num[p.first][p.second] == 1) vse[p.first * k + p.second] = 0;
        nx[p.first][p.second] |= vse;
        if (num[p.first][p.second] == 1) vse[p.first * k + p.second] = 1;
      }
      for (auto p : I[i]) {
        num[p.first][p.second]--;
      }
    }
    conds.push_back(I[i]);
    for (auto x : I[i]) {
      has[x.first][x.second]++;
    }
  }
  brute({});
  cout << answ << "\n";
  return 0;
}
