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 = (int)2e5 + 5;
vector<int> comp[10][10];
vector<pair<int, int> > adj[MAXN];
vector<int> par[MAXN], ord[MAXN];
vector<pair<int, int> > ehh[10][10];
bool saw[10][10];
int perm[MAXN];
int mark[MAXN];
int deg[MAXN];
int n, m, k, ans;
void rec(int p, int good = 0) {
  if (p == k + 1) {
    ans += (good == n);
    return;
  }
  for (int i = 0; i < p; ++i) {
    bool can = 1;
    for (auto it : ehh[p][i]) {
      if (perm[it.first] == it.second) {
        can = 0;
        break;
      }
    }
    if (can) {
      good += comp[p][i].size();
      perm[p] = i;
      rec(p + 1, good);
      good -= comp[p][i].size();
    }
  }
}
void solve() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 1; i <= m; ++i) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    adj[u].emplace_back(w, v);
    par[v].push_back(u);
    ++deg[u];
  }
  for (int i = 1; i <= n; ++i) {
    sort((adj[i]).begin(), (adj[i]).end());
    for (auto it : adj[i]) {
      ord[i].push_back(it.second);
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (auto j : par[i]) {
      saw[deg[j]][find((ord[j]).begin(), (ord[j]).end(), i) - ord[j].begin()] =
          1;
    }
    vector<pair<int, int> > kek;
    for (int a = 1; a <= k; ++a) {
      vector<pair<int, int> > cur;
      for (int b = 0; b < a; ++b) {
        if (saw[a][b]) {
          cur.emplace_back(a, b);
          saw[a][b] = 0;
          comp[a][b].push_back(i);
        }
      }
      for (pair<int, int> prv : kek) {
        for (pair<int, int> me : cur) {
          ehh[me.first][me.second].push_back(prv);
        }
      }
      for (pair<int, int> me : cur) {
        kek.push_back(me);
      }
    }
  }
  rec(1);
  printf("%d\n", ans);
}
int main() {
  int tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}
