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 nax = 2e5 + 5;
vector<pair<int, int> > adj[nax];
vector<int> inc[nax];
int n, m, k;
int cnt[11];
vector<vector<int> > base;
bool cmp(pair<int, int> x, pair<int, int> y) { return x.second < y.second; }
bool ok[11][11][11][11];
int have[11][11];
long long int go[11][11];
long long int fast[nax];
const int mod = 1e9 + 9;
void rec(int dep, int akt, vector<int> p) {
  p.push_back(akt);
  if (dep == k) {
    base.push_back(p);
    return;
  }
  for (int i = 1; i <= dep + 1; i++) {
    rec(dep + 1, i, p);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  fast[0] = 1;
  for (int i = 1; i < nax; i++) {
    fast[i] = fast[i - 1] * 67;
    fast[i] %= mod;
  }
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    int w;
    cin >> w;
    adj[x].push_back(make_pair(y, w));
    inc[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end(), cmp);
  for (int i = 1; i <= n; i++) {
    vector<pair<int, int> > akt;
    for (int s : inc[i]) {
      for (int j = 0; j < adj[s].size(); j++) {
        if (adj[s][j].first == i) {
          akt.push_back(make_pair(adj[s].size(), j + 1));
        }
      }
    }
    int c = akt.size();
    for (int a = 1; a <= c; a++) {
      pair<int, int> v = akt[a - 1];
      go[v.first][v.second] += fast[i];
    }
  }
  long long int w = 0;
  for (int i = 1; i <= n; i++) {
    w += fast[i];
    w %= mod;
  }
  int ans = 0;
  vector<int> e = {};
  rec(1, 1, e);
  for (auto s : base) {
    vector<int> akt = s;
    long long int sum = 0;
    for (int i = 1; i <= k; i++) {
      sum += go[i][akt[i - 1]];
      sum %= mod;
    }
    if (sum == w) ans++;
  }
  cout << ans;
  return 0;
}
