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;
const int maxn = 200200;
int n, k, m;
vector<pair<int, int> > g[maxn];
set<pair<int, int> > st[maxn];
int ans;
bool bad[12][12];
bool dab[12][12][12][12];
int c[12];
void addin(int v, int z, int f) {
  if (st[v].count({z, f})) {
    bad[z][f] = true;
  } else {
    for (auto it : st[v])
      dab[it.first][it.second][z][f] = dab[z][f][it.first][it.second] = true;
    st[v].insert({z, f});
  }
}
void bt(int i) {
  if (i > k) {
    ans++;
    return;
  }
  for (int d = 0; d < i; d++) {
    if (bad[i][d]) continue;
    bool bd = false;
    for (int j = 1; j < i; j++)
      if (dab[j][c[j]][i][d]) {
        bd = true;
        break;
      }
    if (bd) continue;
    c[i] = d;
    bt(i + 1);
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int v, u, w;
    cin >> v >> u >> w;
    v--, u--;
    g[v].push_back({w, u});
  }
  for (int v = 0; v < n; v++) {
    if (!g[v].size()) {
      cout << 0 << endl;
      return 0;
    }
    sort(g[v].begin(), g[v].end());
    for (int i = 0; i < g[v].size(); i++) addin(g[v][i].second, g[v].size(), i);
  }
  bt(1);
  cout << ans << endl;
  return 0;
}
