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 N = 2e5 + 5;
vector<pair<int, int>> G[N];
set<pair<int, int>> in_edges[N];
set<pair<int, int>> forbidden_pos;
set<pair<pair<int, int>, pair<int, int>>> forbidden_pairs;
int count(vector<int>& C, int k);
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  while (m--) {
    int u, v, w;
    scanf("%d%d%d", &u, &v, &w);
    G[u].emplace_back(w, v);
  }
  for (int i = 1; i <= n; i++) sort(G[i].begin(), G[i].end());
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < int(G[i].size()); j++) {
      if (in_edges[G[i][j].second].count({G[i].size(), j + 1}) != 0)
        forbidden_pos.insert({G[i].size(), j + 1});
      in_edges[G[i][j].second].insert({G[i].size(), j + 1});
    }
  }
  for (int i = 1; i <= n; i++) {
    vector<pair<int, int>> V(in_edges[i].begin(), in_edges[i].end());
    for (auto a : V) {
      for (auto b : V) {
        forbidden_pairs.insert({a, b});
      }
    }
  }
  vector<int> C;
  printf("%d\n", count(C, k));
}
int count(vector<int>& C, int k) {
  if (int(C.size()) == k) {
    return 1;
  }
  int res = 0;
  for (int i = 1; i <= int(C.size() + 1); i++) {
    if (forbidden_pos.count({C.size() + 1, i}) != 0) continue;
    bool conflict = false;
    for (int j = 0; j < int(C.size()); j++) {
      if (forbidden_pairs.count({{j + 1, C[j]}, {C.size() + 1, i}}) != 0) {
        conflict = true;
        break;
      }
    }
    if (!conflict) {
      C.push_back(i);
      res += count(C, k);
      C.pop_back();
    }
  }
  return res;
}
