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;
int n, m, k, eu, ev, ew;
vector<pair<int, int> > ga[200100];
struct set_hash {
  long long hs[3], mo[3];
  set_hash() {
    mo[0] = 998244353;
    mo[1] = 1000000007;
    mo[2] = 13331;
    hs[0] = 0;
    hs[1] = 1;
    hs[2] = 0;
  }
  set_hash(long long v) {
    mo[0] = 998244353;
    mo[1] = 1000000007;
    mo[2] = 13331;
    hs[0] = v;
    hs[1] = v;
    hs[2] = v;
  }
  set_hash operator+(set_hash v) {
    set_hash w;
    w.hs[0] = (hs[0] + v.hs[0]) % mo[0];
    w.hs[1] = (hs[1] * v.hs[1]) % mo[1];
    w.hs[2] = (hs[2] + v.hs[2]) % mo[2];
    return w;
  }
  void operator+=(set_hash v) {
    hs[0] = (hs[0] + v.hs[0]) % mo[0];
    hs[1] = (hs[1] * v.hs[1]) % mo[1];
    hs[2] = (hs[2] + v.hs[2]) % mo[2];
  }
};
bool operator==(set_hash a, set_hash b) {
  return ((a.hs[0] == b.hs[0]) && (a.hs[1] == b.hs[1]) && (a.hs[2] == b.hs[2]));
}
set_hash val[10][10], tot;
bool vis[10];
int res;
void bao(int pos, set_hash ha) {
  if (pos == k + 1) {
    if (ha == tot) res++;
    return;
  }
  set_hash nx;
  for (int i = 0; i < pos; i++) {
    nx = ha + val[pos][i];
    bao(pos + 1, nx);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m >> k;
  while (m--) {
    cin >> eu >> ev >> ew;
    ga[eu].push_back(make_pair(ew, ev));
  }
  for (int i = 1; i <= n; i++) sort(ga[i].begin(), ga[i].end());
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < ga[i].size(); j++)
      val[ga[i].size()][j] += ga[i][j].second;
  for (int i = 1; i <= n; i++) tot += i;
  bao(1, set_hash());
  cout << res << endl;
  return 0;
}
