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;
long long ex[10][10] = {};
long long msk(int ck, int p) {
  int idx = (ck <= 4) ? (ck - 1) * 4 + p : 16 + (ck - 5) * 9 + p;
  return 1ll << idx;
}
bool has[10] = {};
long long mask;
int path[10];
int k;
int res;
void calc(int ck) {
  if (ck > k) {
    res++;
    return;
  }
  if (!has[ck]) {
    calc(ck + 1);
    return;
  }
  for (int i = 0; i < ck; i++) {
    if ((mask & msk(ck, i)) != 0) continue;
    path[ck] = i;
    long long em = ex[ck][i];
    int p = ck;
    while (p >= 1 && (!has[p] || ((em & msk(p, path[p])) == 0))) p--;
    if (p >= 1) continue;
    long long mc = mask;
    mask |= em;
    calc(ck + 1);
    mask = mc;
  }
}
int main() {
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m >> k;
  vector<vector<array<int, 2>>> out(n + 1);
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    out[u].push_back({w, v});
  }
  vector<vector<array<int, 2>>> in(n + 1);
  for (int i = 1; i <= n; i++) {
    sort(out[i].begin(), out[i].end());
    has[out[i].size()] = true;
    for (int j = 0; j < out[i].size(); j++) {
      int p = out[i][j][1];
      in[p].push_back({(int)out[i].size(), j});
    }
  }
  for (int i = 1; i <= n; i++) {
    sort(in[i].begin(), in[i].end());
    auto& v = in[i];
    int p = 0;
    long long m1 = 0;
    long long m2 = 0;
    while (p < v.size()) {
      int ps = p;
      m1 |= msk(v[p][0], v[p][1]);
      p++;
      while (p < v.size() && v[p][0] == v[ps][0]) {
        m1 |= msk(v[p][0], v[p][1]);
        if (v[p][1] == v[p - 1][1]) m2 |= msk(v[p][0], v[p][1]);
        p++;
      }
    }
    for (int ck = 1; ck <= k; ck++) {
      for (int i = 0; i < ck; i++) {
        long long sm = msk(ck, i);
        if ((m1 & sm) != 0) {
          if ((m2 & sm) != 0) {
            ex[ck][i] |= m1;
          } else {
            ex[ck][i] |= (m1 & ~sm);
          }
        }
      }
    }
  }
  long long mul = 1;
  for (int i = 1; i <= k; i++) {
    if (!has[i]) mul *= i;
  }
  mask = 0;
  res = 0;
  calc(1);
  cout << res * mul << endl;
  return 0;
}
