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;
inline long long getint() {
  char c = getchar();
  long long ret = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') ret = ret * 10 + c - '0', c = getchar();
  return ret * f;
}
const int maxn = 2e5 + 10;
vector<int> P[10][10];
vector<pair<int, int> > E[maxn];
int n, m, k, ans, deg[maxn];
int na[110][110], nc[110], laji[110];
int sta[110];
inline void dfs(int dep) {
  if (dep > k) {
    ans++;
    return;
  }
  for (int i = 1; i <= dep; i++) {
    int p = (dep - 1) * k + i;
    if (laji[p]) continue;
    if (nc[p]) continue;
    for (int q = 1; q <= k * k; q++)
      if (na[p][q]) nc[q]++;
    sta[dep] = i;
    dfs(dep + 1);
    for (int q = 1; q <= k * k; q++)
      if (na[p][q]) nc[q]--;
  }
}
int main() {
  n = getint();
  m = getint();
  k = getint();
  for (int i = 1; i <= m; i++) {
    int u = getint(), v = getint(), w = getint();
    E[u].push_back(make_pair(w, v));
    deg[u]++;
  }
  for (int i = 1; i <= n; i++) sort(E[i].begin(), E[i].end());
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= deg[i]; j++)
      P[deg[i]][j].push_back(E[i][j - 1].second);
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= i; j++) sort(P[i][j].begin(), P[i][j].end());
  for (int i = 1; i <= k; i++)
    for (int j = i + 1; j <= k; j++)
      for (int p = 1; p <= i; p++)
        for (int q = 1; q <= j; q++) {
          int u = (i - 1) * k + p, v = (j - 1) * k + q;
          int now1 = 0, now2 = 0;
          for (auto now1 : P[i][p]) {
            while (now2 < P[j][q].size() && P[j][q][now2] < now1) now2++;
            if (now2 == P[j][q].size()) break;
            if (now1 == P[j][q][now2]) {
              na[u][v] = 1;
              break;
            }
          }
        }
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= i; j++) {
      int u = (i - 1) * k + j, flag = 0;
      for (int p = 0; p < (int)P[i][j].size() - 1; p++)
        if (P[i][j][p] == P[i][j][p + 1]) laji[u] = 1;
    }
  for (int i = 1; i <= k * k; i++)
    for (int j = 1; j < i; j++) na[i][j] = na[j][i];
  dfs(1);
  printf("%d\n", ans);
  return 0;
}
