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 ull = unsigned long long;
mt19937 mt(736);
bool nxt(vector<int> &arr) {
  for (int i = 0; i < arr.size(); i++)
    if (arr[i]++ == i)
      arr[i] = 0;
    else
      return true;
  return false;
}
void solve(istream &cin = std::cin, ostream &cout = std::cout) {
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<pair<int, int>>> gr(n);
  for (int i = 0; i < m; i++) {
    int a, b, w;
    cin >> a >> b >> w;
    a--;
    b--;
    gr[a].emplace_back(w, b);
  }
  for (auto &row : gr) sort(row.begin(), row.end());
  vector<int> c(k);
  vector<vector<bitset<(int)2e5 + 1>>> q(k);
  for (int i = 0; i < q.size(); i++) {
    q[i].resize(i + 1);
    for (int j = 0; j < q[i].size(); j++)
      for (const auto &vert : gr)
        if (vert.size() - 1 == i) {
          if (q[i][j].test(vert[j].second))
            q[i][j].set(q[i][j].size() - 1);
          else
            q[i][j].set(vert[j].second);
        }
  }
  map<tuple<int, int, int, int>, bool> dp;
  auto ask = [&dp, &q](int a, int b, int c, int d) {
    if (!dp.count({a, b, c, d}))
      dp[{a, b, c, d}] = (q[a][b] & q[c][d]).none() &&
                         !q[a][b].test(q[a][b].size() - 1) &&
                         !q[c][d].test(q[c][d].size() - 1);
    return dp[{a, b, c, d}];
  };
  int ans = 0;
  do {
    bool ok = !q[0][0].test(q[0][0].size() - 1);
    for (int i = 0; ok && i < q.size(); i++)
      for (int j = i + 1; ok && j < q.size(); j++) ok &= ask(i, c[i], j, c[j]);
    if (ok) ans++;
  } while (nxt(c));
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed;
  solve();
  return 0;
}
