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;
int n, m, k;
long long masks[9][9][1563];
bool good[9][9][9][9];
bool selfgood[9][9];
int ans;
int c[9];
void rec(int cur) {
  if (cur == k) {
    ans++;
    return;
  }
  for (int i = 0; i <= cur; i++) {
    bool ok = selfgood[cur][i];
    for (int j = 0; j < cur && ok; j++)
      if (!good[j][c[j]][cur][i]) {
        ok = false;
        break;
      }
    if (ok) {
      c[cur] = i;
      rec(cur + 1);
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  vector<vector<pair<int, int>>> adj(n);
  while (m--) {
    int u, v, w;
    cin >> u >> v >> w;
    u--;
    v--;
    adj[u].emplace_back(w, v);
  }
  for (int i = 0; i < n; i++) sort(adj[i].begin(), adj[i].end());
  memset(masks, 0, sizeof(masks));
  memset(selfgood, true, sizeof(selfgood));
  for (int i = 0; i < n; i++) {
    int d = adj[i].size();
    for (int j = 0; j < d; j++) {
      int v = adj[i][j].second;
      int x = v >> 6;
      int y = v & 63;
      if (masks[d - 1][j][x] & (1LL << y)) selfgood[d - 1][j] = false;
      masks[d - 1][j][x] |= (1LL << y);
    }
  }
  memset(good, true, sizeof(good));
  for (int i = 0; i < k; i++)
    for (int j = 0; j <= i; j++)
      for (int kk = i + 1; kk < k; kk++)
        for (int t = 0; t <= kk; t++) {
          bool ok = true;
          for (int a = 0; a < 1563; a++)
            if (masks[i][j][a] & masks[kk][t][a]) {
              ok = false;
              break;
            }
          if (!ok) {
            good[i][j][kk][t] = good[kk][t][i][j] = false;
          }
        }
  ans = 0;
  rec(0);
  cout << ans << "\n";
  return 0;
}
