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;
constexpr static int MAXN = 2e5 + 10;
int n, m, k;
vector<pair<int, int>> adj[MAXN];
int val[MAXN][10];
bool check[10][10][10][10];
int cnt[10][10];
int c[10];
int calc(int i, int cur, int added) {
  for (int j = 0; j < i; j++)
    if (check[j][c[j]][i][cur]) return 0;
  c[i] = cur;
  added += cnt[i][cur];
  if (i == k - 1) return added == n ? 1 : 0;
  int total = 0;
  for (int j = 0; j <= i + 1; j++) total += calc(i + 1, j, added);
  return total;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m >> k;
  for (int j = 0; j < m; j++) {
    int u, v, w;
    cin >> u >> v >> w;
    u--, v--;
    adj[u].push_back({w, v});
  }
  for (int i = 0; i < n; i++) {
    sort(adj[i].begin(), adj[i].end());
    int l = adj[i].size() - 1;
    for (int p = 0; p < adj[i].size(); p++) {
      int j = adj[i][p].second;
      val[j][l] |= (1 << p);
    }
  }
  for (int i = 0; i < n; i++) {
    vector<vector<int>> idx(k);
    for (int j = 0; j < k; j++)
      for (int v = 0; v < k; v++)
        if (val[i][j] & (1 << v)) {
          idx[j].push_back(v);
          cnt[j][v]++;
        }
    for (int j1 = 0; j1 < k; j1++)
      for (int j2 = j1 + 1; j2 < k; j2++)
        for (int v1 : idx[j1])
          for (int v2 : idx[j2]) check[j1][v1][j2][v2] = true;
  }
  cout << calc(0, 0, 0) << endl;
}
