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 vi = vector<int>;
using ld = long double;
using pii = pair<int, int>;
const int K = 10;
const int N = (int)2e5 + 10;
bool bad[K][K][K][K];
int go(int i, int k, vi &have) {
  if (i == k) return 1;
  int ans = 0;
  for (int j = 0; j < i + 1; j++) {
    bool skip = false;
    have[i] = j;
    for (int q = 0; q < i + 1; q++) skip |= bad[i][j][q][have[q]];
    if (skip) continue;
    ans += go(i + 1, k, have);
  }
  return ans;
}
void solve() {
  for (int i = 0; i < K; i++)
    for (int j = 0; j < K; j++)
      for (int s = 0; s < K; s++)
        for (int t = 0; t < K; t++) bad[i][j][s][t] = false;
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<pii>> adj(n);
  vi deg(n);
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    u--, v--;
    adj[u].emplace_back(w, v);
    deg[u]++;
  }
  for (int i = 0; i < n; i++) sort(begin(adj[i]), end(adj[i]));
  vector<vector<bitset<N>>> eds(k);
  for (int i = 0; i < k; i++)
    for (int j = 0; j < i + 1; j++) eds[i].emplace_back();
  for (int i = 0; i < n; i++)
    for (int j = 0; j < deg[i]; j++) {
      if (eds[deg[i] - 1][j][adj[i][j].second])
        bad[deg[i] - 1][j][deg[i] - 1][j] = true;
      eds[deg[i] - 1][j][adj[i][j].second] = 1;
    }
  for (int i = 0; i < k; i++)
    for (int j = 0; j < i + 1; j++)
      for (int s = 0; s < k; s++)
        for (int t = 0; t < s + 1; t++)
          if (i != s)
            if ((eds[i][j] & eds[s][t]).count()) bad[i][j][s][t] = true;
  vi have(K);
  cout << go(0, k, have) << '\n';
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
  solve();
  return 0;
}
