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>
#pragma GCC optimize("Ofast")
using namespace std;
const int MXN = 2e5 + 10;
const int MXK = 11;
int n, m, k, ans;
int C[MXN];
vector<pair<int, int>> G[MXN];
vector<int> adj[MXN], Ver[MXK];
bitset<MXN> will[MXK][MXK], nw;
bool bad[MXK][MXK], bd[MXK][MXK][MXK][MXK];
void solve(int p) {
  if (p == k + 1) {
    ans++;
    return;
  }
  for (int i = 1; i <= p; i++) {
    if (bad[p][i]) continue;
    bool f = true;
    for (int p2 = 1; p2 < p; p2++) {
      if (bd[p2][p][C[p2]][i]) {
        f = false;
        break;
      }
    }
    if (!f) continue;
    C[p] = i;
    solve(p + 1);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    G[u].push_back({w, v});
  }
  for (int i = 1; i <= n; i++) {
    sort(G[i].begin(), G[i].end());
    Ver[int(G[i].size())].push_back(i);
    for (auto X : G[i]) adj[i].push_back(X.second);
    G[i].clear();
  }
  for (int p = 1; p <= k; p++) {
    for (int i = 1; i <= p; i++) {
      for (int j = 0; j < Ver[p].size(); j++) {
        will[p][i][adj[Ver[p][j]][i - 1]] = 1;
      }
      if (will[p][i].count() < Ver[p].size()) bad[p][i] = 1;
    }
  }
  for (int p1 = 1; p1 <= k; p1++) {
    for (int p2 = p1 + 1; p2 <= k; p2++) {
      for (int i1 = 1; i1 <= p1; i1++) {
        for (int i2 = 1; i2 <= p2; i2++) {
          if (bad[p1][i1] || bad[p2][i2]) continue;
          nw = will[p1][i1] & will[p2][i2];
          if (nw.count()) bd[p1][p2][i1][i2] = 1;
        }
      }
    }
  }
  solve(1);
  cout << ans << '\n';
  return 0;
}
