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;
constexpr ll lim = 10;
bool forbidden[lim][lim][lim][lim];
int state[lim];
ll res = 0;
void check(ll k, ll pos = 1) {
  if (pos > k) {
    for (ll i = 1; i <= k; i++) {
      for (ll j = i; j <= k; j++) {
        if (forbidden[i][state[i]][j][state[j]]) return;
      }
    }
    res++;
  } else {
    for (ll i = 0; i < pos; i++) {
      state[pos] = i;
      check(k, pos + 1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  ll n, m, k;
  cin >> n >> m >> k;
  vector<vector<pair<ll, ll>>> adj(n);
  for (ll i = 0; i < m; i++) {
    ll a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    adj[a].push_back({c, b});
  }
  vector<map<pair<ll, ll>, ll>> rev(n);
  for (ll i = 0; i < n; i++) {
    sort(begin(adj[i]), end(adj[i]));
    for (ll j = 0; j < (ll)(adj[i]).size(); j++) {
      ll to = adj[i][j].second;
      rev[to][{(ll)(adj[i]).size(), j}]++;
    }
  }
  for (ll i = 0; i < n; i++) {
    for (auto e : rev[i]) {
      if (e.second > 1)
        forbidden[e.first.first][e.first.second][e.first.first]
                 [e.first.second] = true;
      for (auto e2 : rev[i]) {
        if (e2.first == e.first) continue;
        forbidden[e.first.first][e.first.second][e2.first.first]
                 [e2.first.second] = true;
      }
    }
  }
  check(k);
  cout << res << endl;
}
