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;
using ll = long long;
int n, m, k;
vector<vector<pair<int, int>>> g;
vector<int> c;
int res;
vector<map<int, int>> enemy;
vector<vector<char>> enemies;
map<pair<int, int>, int> pair_to_idx;
void rec(int idx) {
  if (idx == k + 1) {
    ++res;
  } else {
    for (int i = 1; i <= idx; ++i) {
      pair<int, int> we(idx, i);
      bool good = true;
      if (pair_to_idx.count(we)) {
        good = !enemies[pair_to_idx[we]][pair_to_idx[we]];
        for (int j = 1; j < idx; ++j) {
          pair<int, int> them(j, c[j]);
          if (pair_to_idx.count(them) &&
              enemies[pair_to_idx[we]][pair_to_idx[them]]) {
            good = false;
            break;
          }
        }
      }
      if (good) {
        c[idx] = i;
        rec(idx + 1);
        c[idx] = -1;
      }
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  while (cin >> n >> m >> k) {
    g.assign(n + 1, vector<pair<int, int>>());
    for (int i = 0; i < m; ++i) {
      int u, v, w;
      cin >> u >> v >> w;
      g[u].emplace_back(w, v);
    }
    enemy.assign(n + 1, map<int, int>());
    for (int i = 1; i <= n; ++i) {
      sort(g[i].begin(), g[i].end());
      int sz = g[i].size();
      for (int j = 0; j < sz; ++j) {
        int to = g[i][j].second;
        pair<int, int> pa = {sz, j + 1};
        int pa_idx = -1;
        if (pair_to_idx.count(pa)) {
          pa_idx = pair_to_idx[pa];
        } else {
          pa_idx = pair_to_idx.size();
          pair_to_idx[pa] = pa_idx;
        }
        enemy[to][pa_idx]++;
      }
    }
    enemies.assign(pair_to_idx.size(), vector<char>(pair_to_idx.size()));
    for (int i = 1; i <= n; ++i) {
      for (auto it = enemy[i].begin(); it != enemy[i].end(); ++it) {
        if (it->second > 1) {
          enemies[it->first][it->first] = true;
        }
        for (auto jt = enemy[i].begin(); jt != enemy[i].end(); ++jt) {
          if (it != jt) {
            enemies[it->first][jt->first] = true;
          }
        }
      }
    }
    c.assign(k + 1, -1);
    res = 0;
    rec(1);
    cout << res << "\n";
  }
}
