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;
const int MAXN = 200100;
bool ok[10][9];
bitset<MAXN> results[10][9];
unordered_map<bitset<MAXN>, int> m1;
unordered_map<bitset<MAXN>, int> m2;
bitset<MAXN> curmask;
void recurse(int i, int j) {
  if (i == j) {
    m1[curmask]++;
    return;
  }
  for (int c = 0; c < i; c++) {
    if (ok[i][c]) {
      curmask ^= results[i][c];
      recurse(i + 1, j);
      curmask ^= results[i][c];
    }
  }
}
vector<pair<int, int> > adj[MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    --u;
    --v;
    adj[u].push_back(make_pair(w, v));
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 0; j < i; j++) {
      ok[i][j] = true;
    }
  }
  for (int i = 0; i < m; i++) {
    sort(adj[i].begin(), adj[i].end());
    int deg = adj[i].size();
    for (int j = 0; j < deg; j++) {
      if (results[deg][j].test(adj[i][j].second)) {
        ok[deg][j] = false;
      } else {
        results[deg][j].set(adj[i][j].second);
      }
    }
  }
  curmask.reset();
  int k1 = max(1, k - 3);
  recurse(1, k1);
  curmask.reset();
  swap(m1, m2);
  recurse(k1, k + 1);
  int ans = 0;
  bitset<MAXN> t;
  bitset<MAXN> complete;
  complete.reset();
  for (int i = 0; i < n; i++) {
    complete.set(i);
  }
  for (auto& it : m1) {
    t = it.first;
    t ^= complete;
    auto it2 = m2.find(t);
    if (it2 != m2.end()) {
      ans += it.second * it2->second;
    }
  }
  cout << ans << endl;
}
