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;
using ll = long long;
using pii = pair<int, int>;
ll nope[50];
int never[50];
vector<pii> v[200003];
vector<int> in[200003];
int n, m, k;
int ans = 0;
void dfs(int pos, ll cnope) {
  if (pos > k) {
    ans++;
    return;
  }
  for (int i = 0; i < pos; i++) {
    int x = pos * (pos - 1) / 2 + i;
    if (never[x]) continue;
    if (cnope & (1LL << x)) continue;
    ll nxt_nope = cnope | nope[x];
    dfs(pos + 1, nxt_nope);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int a, b, x;
    cin >> a >> b >> x;
    v[a].push_back({x, b});
  }
  for (int i = 1; i <= n; i++) {
    sort(v[i].begin(), v[i].end());
    int len = v[i].size();
    for (int j = 0; j < v[i].size(); j++) {
      in[v[i][j].second].push_back(len * (len - 1) / 2 + j);
    }
  }
  for (int i = 1; i <= n; i++) {
    sort(in[i].begin(), in[i].end());
    for (int j = 1; j < in[i].size(); j++) {
      if (in[i][j] == in[i][j - 1]) never[in[i][j]] = 1;
    }
    in[i].resize(unique(in[i].begin(), in[i].end()) - in[i].begin());
    for (int x = 0; x < in[i].size(); x++) {
      for (int y = 0; y < x; y++) {
        nope[in[i][x]] |= (1LL << in[i][y]);
        nope[in[i][y]] |= (1LL << in[i][x]);
      }
    }
  }
  dfs(1, 0);
  cout << ans;
}
