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;
void yes() {
  cout << "YES"
       << "\n";
}
void no() {
  cout << "NO"
       << "\n";
}
int dx[] = {1, 0, -1, 0, -1, -1, 1, 1};
int dy[] = {0, 1, 0, -1, -1, 1, -1, 1};
const long long INF = 8e18 + 8;
const double EPS = 1e-9;
const long long MOD = 1e9 + 7;
vector<pair<int, int> > g[200010];
vector<pair<int, int> > in[200010];
set<int> s[10][10];
bool bad[10][10][10][10];
int k;
int ans = 0;
vector<int> v;
void rec(int pos) {
  if (pos == k) {
    for (int i = 0; i < k; i++) {
      for (int j = i + 1; j < k; j++) {
        if (bad[i + 1][v[i]][j + 1][v[j]]) return;
      }
    }
    ans++;
    return;
  }
  for (int i = 1; i <= pos + 1; i++) {
    if (bad[pos + 1][i][pos + 1][i]) continue;
    v.push_back(i);
    rec(pos + 1);
    v.pop_back();
  }
}
void solve() {
  int n, m;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    g[u].push_back(make_pair(w, v));
  }
  for (int i = 1; i <= n; i++) {
    sort((g[i]).begin(), (g[i]).end());
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < (int)((g[i]).size()); j++) {
      int w = g[i][j].first;
      int to = g[i][j].second;
      if (s[(int)((g[i]).size())][j + 1].count(to)) {
        bad[(int)((g[i]).size())][j + 1][(int)((g[i]).size())][j + 1] = true;
      } else {
        s[(int)((g[i]).size())][j + 1].insert(to);
        in[to].push_back(make_pair((int)((g[i]).size()), j + 1));
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < (int)((in[i]).size()); j++) {
      int T = in[i][j].first;
      int t = in[i][j].second;
      for (int q = j + 1; q < (int)((in[i]).size()); q++) {
        int D = in[i][q].first;
        int d = in[i][q].second;
        bad[T][t][D][d] = true;
        bad[D][d][T][t] = true;
      }
    }
  }
  rec(0);
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  bool tests = 0;
  int t = 1;
  if (tests) cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
