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 long long MAX = 3e5 + 5;
long long inf = (long long)30000;
long long mod = (long long)998244353;
long long n, k, m, mb[10][10][10][10];
long long par[MAX];
map<pair<long long, long long>, long long> rev;
vector<pair<long long, long long>> gr[MAX];
vector<pair<long long, long long>> verts[MAX];
set<long long> s[10][10];
long long ans = 0;
vector<long long> cur;
long long pos(long long a, long long b) {
  if (mb[a][b][a][b]) return 1;
  for (int i = 0; i < cur.size(); i++) {
    if (mb[i + 1][cur[i]][a][b]) return 1;
  }
  return 0;
}
void rec() {
  if (cur.size() == k) {
    ans++;
    return;
  } else {
    for (int i = 1; i <= cur.size() + 1; i++) {
      if (!pos(cur.size() + 1, i)) {
        cur.push_back(i);
        rec();
        cur.pop_back();
      }
    }
  }
}
void solve() {
  cin >> n >> m >> k;
  long long cur = 0;
  for (int i = 1; i <= m; i++) {
    long long a, b, w;
    cin >> a >> b >> w;
    gr[a].push_back({w, b});
  }
  for (int i = 1; i <= n; i++) {
    sort(gr[i].begin(), gr[i].end());
    long long len = gr[i].size();
    for (int j = 1; j <= len; j++) {
      verts[gr[i][j - 1].second].push_back({len, j});
      s[len][j].insert(gr[i][j - 1].second);
    }
  }
  for (int i = 1; i <= n; i++) {
    sort(verts[i].begin(), verts[i].end());
    for (int j = 1; j < verts[i].size(); j++) {
      if (verts[i][j] == verts[i][j - 1]) {
        long long a = verts[i][j].first;
        long long b = verts[i][j].second;
        mb[a][b][a][b] = 1;
      }
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      for (int z = 1; z <= k; z++) {
        for (int l = 1; l <= z; l++) {
          if (make_pair(i, j) == make_pair(z, l)) continue;
          if (mb[i][j][z][l]) continue;
          for (auto v : s[i][j]) {
            if (s[z][l].count(v)) {
              mb[i][j][z][l] = 1;
              mb[z][l][i][j] = 1;
              break;
            }
          }
        }
      }
    }
  }
  rec();
  cout << ans;
}
int main() {
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  ios::sync_with_stdio(0);
  long long q;
  q = 1;
  while (q--) solve();
  return 0;
}
