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 MOD = 1e9 + 7;
template <class T>
void chmin(T &a, const T &b) {
  if (a > b) a = b;
}
template <class T>
void chmax(T &a, const T &b) {
  if (a < b) a = b;
}
int N, M, K;
int ng[11][11][11][11];
int ans = 0;
void dfs(int now, vector<pair<int, int>> &v) {
  if (now > K) {
    ans++;
    return;
  }
  for (int j = 1; j <= now; j++) {
    bool ok = true;
    if (ng[now][j][now][j]) continue;
    for (int k = 0; k < v.size(); k++) {
      if (ng[now][j][v[k].first][v[k].second]) ok = false;
    }
    if (ok == false) continue;
    v.push_back(make_pair(now, j));
    dfs(now + 1, v);
    v.pop_back();
  }
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> N >> M >> K;
  vector<vector<pair<int, int>>> g(N);
  for (int i = 0; i < (M); ++i) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    g[a].push_back(make_pair(c, b));
  }
  for (int i = 0; i < (N); ++i) sort(g[i].begin(), g[i].end());
  for (int i = 0; i < (N); ++i)
    for (int j = 0; j < (g[i].size()); ++j) swap(g[i][j].first, g[i][j].second);
  vector<vector<int>> nodes(K + 1);
  for (int i = 0; i < (N); ++i) nodes[g[i].size()].push_back(i);
  vector<vector<pair<int, int>>> in(N);
  for (int i = 1; i <= K; i++) {
    for (int j = 1; j <= i; j++) {
      for (auto now : nodes[i]) {
        int b = g[now][j - 1].first;
        in[b].push_back(make_pair(i, j));
      }
    }
  }
  for (int i = 0; i < N; i++) {
    map<pair<int, int>, int> mp;
    for (auto p : in[i]) {
      if (mp.count(p)) {
        int a = p.first;
        int b = p.second;
        ng[a][b][a][b] = 1;
      }
      mp[p] = 1;
    }
    sort(in[i].begin(), in[i].end());
    in[i].erase(unique(in[i].begin(), in[i].end()), in[i].end());
    for (int j = 0; j < in[i].size(); j++)
      for (int k = 0; k < j; k++) {
        ng[in[i][j].first][in[i][j].second][in[i][k].first][in[i][k].second] =
            1;
        ng[in[i][k].first][in[i][k].second][in[i][j].first][in[i][j].second] =
            1;
      }
  }
  vector<pair<int, int>> v;
  dfs(1, v);
  cout << ans << endl;
  return 0;
}
