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 long long INF = 1e18;
const long long MOD = 998244353;
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) int idx,
               __attribute__((unused)) int LINE_NUM) {
  cerr << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
  if (idx > 0)
    cerr << ", ";
  else
    cerr << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cerr << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
bool have[11][11][200010];
bool illegal[12][12][12][12];
int n, m, k;
int ans = 0;
void check(vector<int> &arr, int pos) {
  if (pos == k) {
    bool state = true;
    for (int i = 0; i < k and state; i++) {
      for (int j = i; j < k and state; j++) {
        if (illegal[i][arr[i]][j][arr[j]]) state = false;
      }
    }
    ans += state;
    return;
  }
  for (int i = 0; i <= pos; i++) {
    arr[pos] = i;
    check(arr, pos + 1);
  }
}
int32_t main() {
  ios::sync_with_stdio(!cin.tie(NULL));
  cin >> n >> m >> k;
  memset(have, false, sizeof(have));
  memset(illegal, false, sizeof(illegal));
  vector<vector<pair<int, int>>> edges(n + 1);
  int u, v, w;
  for (int i = 0; i < m; i++) {
    cin >> u >> v >> w;
    edges[u].push_back({w, v});
  }
  for (int i = 1; i <= n; i++) {
    if (edges[i].empty()) {
      cout << 0;
      return 0;
    }
    sort(edges[i].begin(), edges[i].end());
    for (int j = 0; j < (int)edges[i].size(); j++) {
      if (have[(int)edges[i].size() - 1][j][edges[i][j].second])
        illegal[(int)edges[i].size() - 1][j][edges[i].size() - 1][j] = true, 42;
      have[(int)edges[i].size() - 1][j][edges[i][j].second] = true;
      42;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int a = 0; a < k; a++) {
      for (int b = 0; b <= a; b++) {
        if (!have[a][b][i]) continue;
        for (int c = 0; c < k; c++) {
          for (int d = 0; d <= c; d++) {
            if (a != c or b != d) illegal[a][b][c][d] |= have[c][d][i];
            if (have[c][d][i]) 42;
          }
        }
      }
    }
  }
  vector<int> arr(k);
  check(arr, 0);
  cout << ans;
}
