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 pii = pair<int, int>;
const int K = 10;
bool two[K][K][K][K];
bool one[K][K];
int n, m, k;
int Go(vector<int>& sofar, int pos) {
  if (pos == k + 1) {
    return 1;
  }
  int ans = 0;
  for (int i = 1; i <= pos; ++i) {
    if (one[i][pos]) continue;
    bool stop = false;
    for (int k = 0; k < sofar.size(); ++k) {
      if (two[sofar[k]][k + 1][i][pos]) {
        stop = true;
        break;
      }
    }
    if (!stop) {
      sofar.push_back(i);
      ans += Go(sofar, pos + 1);
      sofar.pop_back();
    }
  }
  return ans;
}
void Run() {
  cin >> n >> m >> k;
  vector<vector<pii>> out(n, vector<pii>{});
  vector<set<pii>> in(n, set<pii>{});
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    --u, --v;
    out[u].emplace_back(w, v);
  }
  for (int i = 0; i < n; ++i) {
    sort(out[i].begin(), out[i].end());
    for (int j = 0; j < out[i].size(); ++j) {
      out[i][j].first = j + 1;
    }
    for (const auto& [x, y] : out[i]) {
      auto res = in[y].emplace(x, out[i].size());
      if (!res.second) {
        one[x][out[i].size()] = true;
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    vector<pii> v(in[i].begin(), in[i].end());
    for (int j = 0; j < v.size(); ++j) {
      for (int k = 0; k < j; ++k) {
        two[v[j].first][v[j].second][v[k].first][v[k].second] = true;
        two[v[k].first][v[k].second][v[j].first][v[j].second] = true;
      }
    }
  }
  vector<int> v;
  cout << Go(v, 1) << '\n';
}
int main() {
  srand(time(0));
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  Run();
  return 0;
}
