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> V[10][10];
vector<pair<int, int>> adj[200000];
vector<int> temp;
int ans, K, CC[10][10];
long long B[200000];
bool imp[10][10], R[50][50];
void solve(int c) {
  if (c > K) {
    ans++;
    return;
  }
  for (int i = 0; i < c; i++)
    if (!imp[c][i]) {
      bool valid = true;
      for (auto t : temp)
        if (!R[t][CC[c][i]]) {
          valid = false;
          break;
        }
      if (valid) {
        temp.push_back(CC[c][i]);
        solve(c + 1);
        temp.pop_back();
      }
    }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ((void)0);
  ((void)0);
  ((void)0);
  int N, M, n = 0;
  cin >> N >> M >> K;
  for (int i = 1; i <= K; i++)
    for (int j = 0; j < i; j++) CC[i][j] = n++;
  while (M--) {
    int a, b, c;
    cin >> a >> b >> c;
    adj[--a].emplace_back(c, --b);
  }
  for (int i = 0; i < N; i++) {
    sort(adj[i].begin(), adj[i].end());
    for (int j = 0; j < adj[i].size(); j++)
      if (!imp[adj[i].size()][j]) {
        if (B[adj[i][j].second] & (1LL << CC[adj[i].size()][j]))
          imp[adj[i].size()][j] = true;
        else {
          V[adj[i].size()][j][adj[i][j].second] = true;
          B[adj[i][j].second] |= 1LL << CC[adj[i].size()][j];
        }
      }
  }
  for (int i = 2; i <= K; i++)
    for (int j = 0; j < i; j++)
      if (!imp[i][j])
        for (int k = 1; k < i; k++)
          for (int l = 0; l < k; l++)
            if (!imp[k][l] && (V[i][j] & V[k][l]).none()) {
              R[CC[i][j]][CC[k][l]] = true;
              R[CC[k][l]][CC[i][j]] = true;
            }
  solve(1);
  cout << ans << '\n';
  return 0;
}
