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 int N = 2e5 + 9, Mod = 1e9 + 9;
int n, m, k, Cnt, Pw[N], H[11][11];
vector<int> Adj[N], V[11];
int SM = 0, targ;
void Solve(int lvl) {
  if (lvl == k + 1) {
    if (SM == targ) Cnt++;
    return;
  }
  for (int j = 0; j < lvl; j++) {
    SM = (SM + H[lvl][j]) % Mod;
    Solve(lvl + 1);
    SM = (SM - H[lvl][j] + Mod) % Mod;
  }
}
int main() {
  for (int i = Pw[0] = 1; i < N; i++) Pw[i] = Pw[i - 1] * 718301903LL % Mod;
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  vector<tuple<int, int, int> > E;
  for (int i = 1; i <= n; i++) targ = (targ + Pw[i]) % Mod;
  for (int i = 0; i < m; i++) {
    int v, u, w;
    cin >> v >> u >> w;
    E.push_back({w, v, u});
  }
  sort(E.begin(), E.end());
  for (auto X : E) {
    int w, v, u;
    tie(w, v, u) = X;
    Adj[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) V[(int)Adj[i].size()].push_back(i);
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < i; j++) {
      int hsh = 0;
      for (int v : V[i]) hsh = (hsh + Pw[Adj[v][j]]) % Mod;
      H[i][j] = hsh;
    }
  }
  Solve(1);
  printf("%d\n", Cnt);
  return 0;
}
