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 vi = vector<int>;
using pii = pair<int, int>;
bitset<200000> bs[9][9];
int allowed[9][9][9];
ll n, m, k;
ll solve(int pos, int val, array<int, 9> still) {
  ll sum = 0;
  for (int i = (pos + 1); i < (k); ++i) {
    still[i] &= allowed[pos][val][i];
    if (still[i] == 0) return 0;
  }
  if (pos == k - 2) {
    return __builtin_popcount(still[k - 1]);
  }
  int del = 0;
  for (int mask = still[pos + 1]; mask; mask &= ~del) {
    del = mask & (-mask);
    sum += solve(pos + 1, __builtin_ctz(del), still);
  }
  return sum;
}
int main(int argc, char* argv[]) {
  std::ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.precision(10);
  cin >> n >> m >> k;
  array<int, 9> still = {};
  for (int i = (0); i < (k); ++i) still[i] = (1u << (i + 1)) - 1;
  vector<vector<pii>> g(n);
  for (int i = (0); i < (m); ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    u--, v--;
    g[u].push_back({w, v});
  }
  for (auto& a : g) sort((a).begin(), (a).end());
  for (int u = (0); u < (n); ++u) {
    for (int i = (0); i < (g[u].size()); ++i) {
      if (bs[g[u].size() - 1][i][g[u][i].second]) {
        still[g[u].size() - 1] &= ~(1u << i);
      }
      bs[g[u].size() - 1][i][g[u][i].second] = true;
    }
  }
  for (int d1 = (0); d1 < (k); ++d1)
    for (int v1 = (0); v1 < (d1 + 1); ++v1)
      for (int d2 = (d1 + 1); d2 < (k); ++d2) {
        allowed[d1][v1][d2] = (1u << (d2 + 1)) - 1;
        for (int v2 = (0); v2 < (d2 + 1); ++v2) {
          if ((bs[d1][v1] & bs[d2][v2]).any()) {
            allowed[d1][v1][d2] &= ~(1u << v2);
          }
        }
      }
  if (k == 1)
    cout << still[0] << '\n';
  else
    cout << (still[0] ? solve(0, 0, still) : 0ll) << '\n';
  cout.flush();
  return 0;
}
