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;
mt19937 gen(time(NULL));
struct Edge {
  int to;
  int weight;
  Edge(const int to_, const int weight_) : to(to_), weight(weight_) {}
};
bool operator<(const Edge& left, const Edge& right) {
  return left.weight < right.weight;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<Edge>> adj(n);
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    adj[u - 1].push_back(Edge(v - 1, w));
  }
  vector<int> outDegree(n);
  for (int i = 0; i < n; i++) {
    outDegree[i] = (int)(adj[i].size());
    sort(adj[i].begin(), adj[i].end());
  }
  vector<long long> hashs(n);
  int total = 0;
  for (int i = 0; i < n; i++) {
    hashs[i] = uniform_int_distribution<int>(1, INT_MAX)(gen);
    total ^= hashs[i];
  }
  vector<vector<long long>> subSets(k + 1, vector<long long>(k + 1));
  for (int i = 0; i < n; i++)
    for (int j = 0; j < outDegree[i]; j++)
      subSets[outDegree[i]][j] ^= hashs[adj[i][j].to];
  int res = 0;
  function<void(int, int)> go = [&](long long curHash, int curDeg) {
    if (curDeg > k) {
      if (curHash == total) res++;
      return;
    }
    for (int nxt = 0; nxt < curDeg; nxt++)
      go(curHash ^ subSets[curDeg][nxt], curDeg + 1);
  };
  go(0LL, 1);
  cout << res << "\n";
  return 0;
}
