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 + 100, K = 20;
vector<pair<int, int> > g[N];
vector<int> r[N];
map<pair<int, int>, int> mp;
bool ok[K][K][K][K];
int c[K];
int n, m, k;
bool val() {
  for (int i = 0; i < k; i++) {
    if (!ok[i][c[i]][i][c[i]]) return false;
    for (int j = i + 1; j < k; j++)
      if (!ok[i][c[i]][j][c[j]]) return false;
  }
  return true;
}
bool nxt() {
  for (int i = 0; i < k; i++)
    if (c[i] < i) {
      c[i]++;
      for (int j = 0; j < i; j++) c[j] = 0;
      return true;
    }
  return false;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m >> k;
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= k; j++)
      for (int l = 0; l <= k; l++)
        for (int h = 0; h <= k; h++) ok[i][j][l][h] = true;
  for (int i = 0; i < m; i++) {
    int v, u, w;
    cin >> v >> u >> w;
    v--;
    u--;
    g[v].push_back(pair<int, int>(w, u));
    r[u].push_back(v);
  }
  for (int v = 0; v < n; v++) {
    sort(g[v].begin(), g[v].end());
    int ind = 0;
    for (auto U : g[v]) mp[pair<int, int>(U.second, v)] = ind++;
  }
  for (int v = 0; v < n; v++) {
    vector<pair<int, int> > vec;
    if (r[v].empty()) {
      cout << "0\n";
      return 0;
    }
    for (auto u : r[v]) {
      int cur = mp[pair<int, int>(v, u)];
      int sz = g[u].size();
      sz--;
      vec.push_back(pair<int, int>(sz, cur));
    }
    sort(vec.begin(), vec.end());
    for (int i = 0; i < vec.size() - 1; i++)
      if (vec[i] == vec[i + 1]) {
        pair<int, int> A = vec[i];
        pair<int, int> B = vec[i + 1];
        ok[A.first][A.second][B.first][B.second] = false;
      }
    vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
    for (int i = 0; i < vec.size(); i++)
      for (auto j = 0; j < vec.size(); j++) {
        if (i == j) continue;
        pair<int, int> A = vec[i];
        pair<int, int> B = vec[j];
        ok[A.first][A.second][B.first][B.second] = false;
      }
  }
  for (int i = 0; i < k; i++) c[i] = 0;
  int ans = 0;
  while (true) {
    ans += val();
    if (!nxt()) break;
  }
  cout << ans << "\n";
  return 0;
}
