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 + 5;
const int C = 10 + 2;
const int INF = 0x3f3f3f3f;
const long long INF_LL = 0x3f3f3f3f3f3f3f3f;
int n, m, c;
struct Adj {
  int v, w;
};
vector<Adj> a[N];
bool ban[C][C][C][C];
int ans;
int arr[N];
void DFS(int dep) {
  if (dep > c) {
    ans++;
    return;
  }
  for (int i = 1; i <= dep; i++) {
    if (ban[dep][i][dep][i]) continue;
    int flag = 1;
    for (int j = 1; j < dep; j++)
      if (ban[dep][i][j][arr[j]]) {
        flag = 0;
        break;
      }
    if (!flag) continue;
    arr[dep] = i;
    DFS(dep + 1);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> m >> c;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    a[u].push_back((Adj){v, w});
  }
  static bool vst[N][C][C];
  for (int i = 1; i <= n; i++) {
    sort(a[i].begin(), a[i].end(), [](Adj a, Adj b) { return a.w < b.w; });
    int siz = a[i].size();
    for (int j = 0; j < siz; j++) {
      auto e = a[i][j];
      if (vst[e.v][siz][j + 1]) ban[siz][j + 1][siz][j + 1] = 1;
      for (int k = 1; k <= c; k++)
        for (int l = 1; l <= k; l++)
          if (vst[e.v][k][l]) {
            ban[siz][j + 1][k][l] = 1;
            ban[k][l][siz][j + 1] = 1;
          }
      vst[e.v][siz][j + 1] = 1;
    }
  }
  DFS(1);
  cout << ans << endl;
  return 0;
}
