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;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class T>
void IN(T& x) {
  cin >> x;
}
template <class H, class... T>
void IN(H& h, T&... t) {
  IN(h);
  IN(t...);
}
template <class T1, class T2>
void OUT(const pair<T1, T2>& x);
template <class T>
void OUT(const T& x) {
  cout << x;
}
template <class H, class... T>
void OUT(const H& h, const T&... t) {
  OUT(h);
  OUT(t...);
}
template <class T1, class T2>
void OUT(const pair<T1, T2>& x) {
  OUT(x.first, ' ', x.second);
}
template <class... T>
void OUTL(const T&... t) {
  OUT(t..., "\n");
}
template <class H>
void OUTLS(const H& h) {
  OUTL(h);
}
template <class H, class... T>
void OUTLS(const H& h, const T&... t) {
  OUT(h, ' ');
  OUTLS(t...);
}
void program();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  program();
}
const int MX = 5e5;
int n, m, k;
vector<pair<int, int> > adj[MX];
long long h[MX];
long long f[10][10];
long long tot;
int getAns(int i, long long bm) {
  if (i == k) {
    if (bm == tot) return 1;
    return 0;
  }
  int ans = 0;
  for (int j = int(0); j < int(i + 1); j++) {
    bm ^= f[i][j];
    ans += getAns(i + 1, bm);
    bm ^= f[i][j];
  }
  return ans;
}
void program() {
  IN(n, m, k);
  for (int i = int(0); i < int(m); i++) {
    int u, v, w;
    IN(u, v, w);
    u--;
    v--;
    adj[u].push_back({w, v});
  }
  for (int i = int(0); i < int(n); i++) sort(adj[i].begin(), adj[i].end());
  for (int i = int(0); i < int(n); i++) h[i] = rng();
  for (int i = int(0); i < int(10); i++)
    for (int j = int(0); j < int(10); j++) f[i][j] = 0;
  for (int u = int(0); u < int(n); u++) {
    for (int i = int(0); i < int(adj[u].size()); i++) {
      f[adj[u].size() - 1][i] ^= h[adj[u][i].second];
    }
  }
  tot = 0;
  for (int i = int(0); i < int(n); i++) tot ^= h[i];
  OUTL(getAns(0, 0));
}
