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;
using ll = long long;
using ld = long double;
using vi = vector<int>;
using vl = vector<ll>;
using pii = pair<int, int>;
using vpii = vector<pair<int, int>>;
template <typename T>
using pque = priority_queue<T, vector<T>, greater<T>>;
constexpr int N = 212345, M = 16;
class Node {
 public:
  int a = 0;
  Node *c[2] = {NULL, NULL};
  Node *next(int a) {
    if (c[a] == NULL) c[a] = new Node();
    return c[a];
  }
} *head = new Node();
vpii g[N];
vi v[M][M];
int p[N] = {};
template <class T>
void dfs(int a, int b, T func) {
  if (a > b) {
    func();
    return;
  };
  for (int i = (0); i < ((a)); i++) {
    for (int j : v[a][i]) p[j]++;
    dfs(a + 1, b, func);
    for (int j : v[a][i]) p[j]--;
  }
}
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = (0); i < ((m)); i++) {
    int a, b, c;
    scanf("%d%d%d", &a, &b, &c);
    g[a].emplace_back(c, b);
  }
  for (int i = (1); i < (n + 1); i++) {
    sort((g[i]).begin(), (g[i]).end());
    for (int j = (0); j < (((int)g[i].size())); j++) {
      v[g[i].size()][j].emplace_back(g[i][j].second);
    }
  }
  for (int i = (1); i < (k); i++)
    for (int j = (0); j < ((i)); j++) {
      ;
      for (int h : v[i][j])
        ;
      ;
    }
  dfs(1, min(6, k), [&]() {
    for (int i = (1); i < (n + 1); i++)
      ;
    ;
    Node *node = head;
    for (int i = (1); i < (n + 1); i++) {
      if (p[i] >= 2) return;
      node = node->next(p[i]);
    }
    node->a++;
  });
  ;
  int ans = 0;
  dfs(min(7, k), k <= 6 ? 0 : k, [&]() {
    Node *node = head;
    for (int i = (1); i < (n + 1); i++) {
      if (p[i] >= 2) return;
      node = node->next(1 - p[i]);
    }
    ans += node->a;
    ;
  });
  printf("%d\n", ans);
}
