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;
int ans, n, m, k, a[12];
struct Edge {
  int to, t;
  Edge() { to = t = 0; }
  Edge(int _to, int _t) {
    to = _to;
    t = _t;
  }
  bool operator<(const Edge &z) const { return t < z.t; }
};
vector<Edge> vec[N];
int d[N][12][12];
bool vis[12][12][12][12];
inline void dfs(int x) {
  if (k + 1 == x) {
    for (int i = 1; i <= k; i++)
      for (int j = 1; j <= k; j++)
        if (vis[i][a[i]][j][a[j]]) return;
    ans++;
    return;
  }
  for (int i = 1; i <= x; i++) a[x] = i, dfs(x + 1);
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  int s, e, t;
  for (int i = 1; i <= m; i++)
    scanf("%d%d%d", &s, &e, &t), vec[s].push_back(Edge(e, t));
  for (int i = 1; i <= n; i++) {
    sort(vec[i].begin(), vec[i].end());
    for (int j = 0; j < vec[i].size(); j++)
      d[vec[i][j].to][vec[i].size()][j + 1]++;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 0; j < vec[i].size(); j++)
      for (int t = 1; t <= k; t++)
        for (int l = 1; l <= k; l++)
          if (d[vec[i][j].to][t][l]) {
            if (t ^ vec[i].size())
              vis[vec[i].size()][j + 1][t][l] = 1;
            else if (d[vec[i][j].to][t][l] > 1)
              vis[vec[i].size()][j + 1][t][l] = 1;
          }
  dfs(1);
  return printf("%d\n", ans), 0;
}
