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;
int hashPair(pair<int, int> pr) { return 10 * pr.first + pr.second; }
int n, m, k;
bool bad[100][100];
vector<int> neighbours[200001];
map<pair<int, int>, int> inEdges[200001];
vector<pair<int, int> > st;
int dfs(int depth) {
  if (depth == k + 1) return 1;
  int ans = 0;
  for (int choice = 0; choice < depth; ++choice) {
    pair<int, int> p = {depth, choice};
    st.push_back(p);
    bool possible = true;
    for (auto &q : st) {
      if (bad[hashPair(p)][hashPair(q)]) {
        possible = false;
        break;
      }
    }
    if (possible) ans += dfs(depth + 1);
    st.pop_back();
  }
  return ans;
}
int main() {
  cin >> n >> m >> k;
  vector<pair<int, int> > edges(m);
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    edges[w - 1] = {u, v};
  }
  for (pair<int, int> edge : edges)
    neighbours[edge.first].push_back(edge.second);
  for (int i = 1; i <= n; ++i) {
    int outDegree = neighbours[i].size();
    for (int j = 0; j < neighbours[i].size(); ++j)
      ++inEdges[neighbours[i][j]][{outDegree, j}];
  }
  for (int i = 1; i <= n; ++i) {
    for (auto a : inEdges[i])
      for (auto b : inEdges[i]) {
        if (a.first != b.first || a.second > 1)
          bad[hashPair(a.first)][hashPair(b.first)] = true;
      }
  }
  cout << dfs(1) << endl;
  return 0;
}
