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;
template <class T>
using vc = vector<T>;
template <class T>
using vvc = vc<vc<T>>;
template <class T>
void mkuni(vector<T>& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
long long rand_int(long long l, long long r) {
  static mt19937_64 gen(chrono::steady_clock::now().time_since_epoch().count());
  return uniform_int_distribution<long long>(l, r)(gen);
}
template <class T>
void print(T x, int suc = 1) {
  cout << x;
  if (suc == 1)
    cout << '\n';
  else
    cout << ' ';
}
template <class T>
void print(const vector<T>& v, int suc = 1) {
  for (int i = 0; i < v.size(); i++)
    print(v[i], i == (int)(v.size()) - 1 ? suc : 2);
}
const int maxn = 2e5 + 7;
struct Edge {
  int to, w;
  bool operator<(const Edge& a) const { return w < a.w; }
};
vc<Edge> G[maxn];
vector<int> deg[10];
int ans = 0;
int n, k;
unsigned long long st[9][9];
unsigned long long val[maxn];
unsigned long long tar;
void solve(int cur, unsigned long long cc) {
  if (cur == k) {
    ans += (cc == tar);
    return;
  }
  for (int i = 0; i <= cur; i++) solve(cur + 1, cc + st[cur][i]);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int m;
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    val[i] = rand_int(0, 5e18);
    tar += val[i];
  }
  for (int i = 0, u, v, w; i < m; i++) {
    cin >> u >> v >> w;
    G[u].push_back(Edge{v, w});
  }
  for (int i = 1; i <= n; i++) {
    sort(G[i].begin(), G[i].end());
    deg[G[i].size() - 1].push_back(i);
  }
  for (int i = 0; i < k; i++) {
    for (int j = 0; j <= i; j++) {
      for (auto u : deg[i]) {
        int to = G[u][j].to;
        st[i][j] += val[to];
      }
    }
  }
  solve(0, 0);
  print(ans);
}
