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;
bitset<200000> choice[10][10];
bool conflict[10][10][10][10];
bool invalid[10][10];
int amt[10][10];
vector<pair<int, int>> adj[200000];
vector<int> by_outdeg[10];
int n, m, k;
int assignment[10];
int ans;
void bt(int i, int sum) {
  if (i == k + 1) {
    ans += (sum == n);
    return;
  }
  for (int j = 1; j <= i; j++) {
    if (invalid[i][j]) continue;
    bool ok = true;
    for (int i2 = 1; i2 < i; i2++)
      if (conflict[i2][assignment[i2]][i][j]) ok = false;
    if (ok) {
      assignment[i] = j;
      bt(i + 1, sum + amt[i][j]);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    adj[u - 1].emplace_back(w, v - 1);
  }
  for (int i = 0; i < n; i++) {
    by_outdeg[adj[i].size()].push_back(i);
    sort(adj[i].begin(), adj[i].end());
  }
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= i; j++) {
      for (int v : by_outdeg[i]) {
        int w = adj[v][j - 1].second;
        if (choice[i][j][w]) {
          invalid[i][j] = true;
          break;
        }
        choice[i][j].set(w);
      }
    }
  for (int i1 = 1; i1 <= k; i1++)
    for (int j1 = 1; j1 <= i1; j1++) {
      amt[i1][j1] = choice[i1][j1].count();
      for (int i2 = i1 + 1; i2 <= k; i2++)
        for (int j2 = 1; j2 <= i2; j2++)
          if ((choice[i1][j1] & choice[i2][j2]).any())
            conflict[i1][j1][i2][j2] = true;
    }
  bt(1, 0);
  cout << ans << endl;
}
