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 = 200010, K = 10;
int n, m, k, res;
vector<pair<int, int>> g[N], gr[N];
bool banned[K][K][K][K];
int perm[K];
void dfs(int pos) {
  if (pos == k) {
    for (int i = 0; i < k; ++i)
      for (int j = i; j < k; ++j)
        if (banned[i][perm[i]][j][perm[j]]) return;
    ++res;
    return;
  }
  for (int i = 0; i <= pos; ++i) {
    perm[pos] = i;
    dfs(pos + 1);
  }
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    --u;
    --v;
    g[u].push_back({w, v});
  }
  for (int i = 0; i < n; ++i) sort(g[i].begin(), g[i].end());
  bool pos = 1;
  for (int i = 0; i < n; ++i) {
    if (!g[i].size()) pos = 0;
    for (int j = 0; j < g[i].size(); ++j)
      gr[g[i][j].second].push_back({g[i].size() - 1, j});
  }
  for (int i = 0; i < n; ++i) {
    bool have[K][K] = {};
    for (int j = 0; j < gr[i].size(); ++j) {
      int a = gr[i][j].first, b = gr[i][j].second;
      if (have[a][b]) banned[a][b][a][b] = 1;
      have[a][b] = 1;
    }
    for (int a = 0; a < k; ++a)
      for (int b = 0; b < k; ++b)
        if (have[a][b])
          for (int c = 0; c < k; ++c)
            for (int d = 0; d < k; ++d)
              if (have[c][d] && (a != c || b != d)) banned[a][b][c][d] = 1;
  }
  dfs(0);
  cout << pos * res << endl;
}
