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;
bool sings[9][9] = {};
bool pairs[9][9][9][9] = {};
int ins[(int)2e5 + 1][9][9] = {};
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<pair<int, int>>> outs(n);
  for (int _ = (0); _ < (m); ++_) {
    int u, v, w;
    cin >> u >> v >> w;
    --u, --v;
    outs[u].emplace_back(w, v);
  }
  for (auto &v : outs) {
    sort((v).begin(), (v).end());
    for (int i = (0); i < (int(v.size())); ++i)
      ins[v[i].second][int(v.size())][i]++;
  }
  for (int v = (0); v < (n); ++v) {
    vector<pair<int, int>> ls;
    for (int s = (1); s < (k + 1); ++s)
      for (int i = (0); i < (s); ++i)
        if (ins[v][s][i] == 1)
          ls.emplace_back(s, i);
        else if (ins[v][s][i] > 1)
          sings[s - 1][i] = true;
    for (int j = (0); j < (int(ls.size())); ++j)
      for (int i = (0); i < (j); ++i)
        pairs[ls[i].first - 1][ls[i].second][ls[j].first - 1][ls[j].second] =
            true;
  }
  int cnt = 0;
  vector<int> vals(k + 1);
  function<void(int)> calc = [&](int s) {
    if (s == k + 1) {
      ++cnt;
      return;
    }
    for (int i = (0); i < (s); ++i)
      if (!sings[s - 1][i]) {
        for (int j = (1); j < (s); ++j)
          if (pairs[j - 1][vals[j]][s - 1][i]) goto foo;
        vals[s] = i;
        calc(s + 1);
      foo:;
      }
  };
  calc(1);
  cout << cnt << endl;
}
