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;
template <typename T>
void chkmax(T &first, T second) {
  first < second ? first = second : T();
}
template <typename T>
void chkmin(T &first, T second) {
  first > second ? first = second : T();
}
template <typename T>
void readint(T &first) {
  first = 0;
  int f = 1;
  char c;
  for (c = getchar(); !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) first = first * 10 + c - '0';
  first *= f;
}
const int MOD = 998244353;
inline int dmy(int first) { return first >= MOD ? first - MOD : first; }
inline void inc(int &first, int second) { first = dmy(first + second); }
int qmi(int first, int second) {
  int ans = 1;
  for (; second; second >>= 1, first = 1ll * first * first % MOD)
    if (second & 1) ans = 1ll * ans * first % MOD;
  return ans;
}
const int MAXN = 200005, MAXK = 11;
int n, k;
vector<pair<int, int> > G[MAXN];
int f[MAXN][MAXK][MAXK];
bool g[MAXK][MAXK][MAXK][MAXK];
int cur[MAXK], ans = 0;
void dfs(int u) {
  if (u > k) {
    ++ans;
    return;
  }
  for (int i = 1; i <= u; ++i) {
    if (g[u][i][u][i]) continue;
    bool flag = 1;
    for (int v = 1; v < u; ++v)
      if (g[u][i][v][cur[v]]) {
        flag = 0;
        break;
      }
    if (!flag) continue;
    cur[u] = i;
    dfs(u + 1);
  }
}
int main() {
  int E, u, v, c;
  readint(n), readint(E), readint(k);
  while (E--) {
    readint(u), readint(v), readint(c);
    G[u].push_back(make_pair(c, v));
  }
  for (int i = 1; i <= n; ++i) {
    sort(G[i].begin(), G[i].end());
    for (int j = 0; j < G[i].size(); ++j)
      f[G[i][j].second][G[i].size()][j + 1]++;
  }
  for (int i = 1; i <= n; ++i)
    for (int first = 1; first <= k; ++first)
      for (int second = 1; second <= first; ++second) {
        if (!f[i][first][second]) continue;
        for (int u = 1; u < first; ++u)
          for (int v = 1; v <= u; ++v)
            if (f[i][u][v]) g[first][second][u][v] = 1;
        if (f[i][first][second] >= 2) g[first][second][first][second] = 1;
      }
  dfs(1);
  cout << ans << endl;
  return 0;
}
