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>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops, no-stack-protector")
#pragma GCC target("avx,avx2,fma")
using namespace std;
using pii = pair<int, int>;
using pll = pair<long long, long long>;
using vb = vector<bool>;
using vi = vector<int>;
using vl = vector<long long>;
using vvb = vector<vector<bool>>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<long long>>;
using vpii = vector<pair<int, int>>;
using mii = map<int, int>;
using umii = unordered_map<int, int>;
using seti = set<int>;
using useti = unordered_set<int>;
template <typename... ArgTypes>
void print(ArgTypes... args);
template <typename... ArgTypes>
void input(ArgTypes &...args);
template <>
void print() {}
template <>
void input() {}
template <typename T, typename... ArgTypes>
void print(T t, ArgTypes... args) {
  cout << t;
  print(args...);
}
template <typename T, typename... ArgTypes>
void input(T &t, ArgTypes &...args) {
  cin >> t;
  input(args...);
}
vb self;
vb visited;
vector<useti> adj2;
int n, m, k;
int traverse(int level) {
  if (level == k) return 1;
  int res = 0;
  vi changed;
  for (int i = 0; i < (level + 1); (i)++) {
    int name = k * level + i;
    if (visited[name] || self[name]) continue;
    for (auto conflict : adj2[name])
      if (!visited[conflict]) {
        changed.push_back(conflict);
        visited[conflict] = 1;
      }
    res += traverse(level + 1);
    for (auto c : changed) visited[c] = 0;
    changed.clear();
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m >> k;
  vector<vpii> adj(n);
  for (int i = 0; i < (m); (i)++) {
    int x, y, w;
    cin >> x >> y >> w;
    x--;
    y--;
    adj[x].push_back({w, y});
  }
  for (int i = 0; i < (n); (i)++) sort((adj[i]).begin(), (adj[i]).end());
  vector<mii> incoming(n);
  for (int i = 0; i < (n); (i)++) {
    int car = ((int)(adj[i]).size());
    for (int j = 0; j < car; j++) {
      int v = adj[i][j].second;
      incoming[v][k * (car - 1) + j]++;
    }
  }
  self = vb(k * k);
  adj2 = vector<useti>(k * k);
  visited = vb(k * k);
  for (int i = 0; i < (n); (i)++) {
    vpii all;
    for (auto d : incoming[i]) all.push_back(d);
    for (int j = 0; j < ((int)(all).size()); j++) {
      if (all[j].second > 1) {
        self[all[j].first] = 1;
        continue;
      }
      for (int k = j + 1; k < ((int)(all).size()); k++)
        adj2[all[j].first].insert(all[k].first);
    }
  }
  cout << traverse(0) << endl;
}
