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;
int n, m, k, ans;
vector<int> tup;
vector<pair<int, int>> g[200000], in[200000];
set<pair<pair<int, int>, pair<int, int>>> bad;
void gen(int x) {
  if (x == k) {
    bool good = true;
    for (int i = 0; i < k; i++)
      for (int j = i; j < k; j++)
        if (bad.count({{i + 1, tup[i]}, {j + 1, tup[j]}})) good = false;
    ans += int(good);
    return;
  }
  for (int i = 0; i < x + 1; i++) {
    tup.push_back(i);
    gen(x + 1);
    tup.pop_back();
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0, u, v, w; i < m; i++) {
    cin >> u >> v >> w;
    g[u - 1].push_back({w, v - 1});
  }
  for (int i = 0; i < n; i++) {
    sort(g[i].begin(), g[i].end());
    for (int j = 0; j < g[i].size(); j++) {
      int cnt = 0;
      pair<int, int> p = {g[i].size(), j};
      for (pair<int, int> x : in[g[i][j].second])
        if (x == p) cnt++;
      if (cnt < 2) in[g[i][j].second].push_back(p);
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < in[i].size(); j++)
      for (int k = j + 1; k < in[i].size(); k++) {
        pair<int, int> t1 = in[i][j], t2 = in[i][k];
        if (t1 > t2) swap(t1, t2);
        bad.insert({t1, t2});
      }
  gen(0);
  cout << ans << endl;
}
