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 int INF = 2e9 + 1;
const int mod = 1e9 + 7;
mt19937 gen(time(0));
int k;
int hashes[10][10];
int ans = 0, need = 0;
void gener(int x, int now) {
  if (x == k + 1) {
    ans += (need == now);
    return;
  }
  for (int y = 1; y <= x; y++) {
    gener(x + 1, (now + hashes[x][y]) % mod);
  }
}
void solve() {
  int n, m;
  cin >> n >> m >> k;
  vector<vector<set<int>>> S(k + 1, vector<set<int>>(k + 1));
  vector<vector<pair<int, int>>> g(n);
  for (int i = 0; i < m; i++) {
    int v, u, w;
    cin >> v >> u >> w;
    v--;
    u--;
    g[v].emplace_back(w, u);
  }
  for (int i = 0; i < n; i++) {
    int deg = g[i].size();
    sort((g[i]).begin(), (g[i]).end());
    for (int j = 1; j <= deg; j++) {
      S[deg][j].insert(g[i][j - 1].second);
    }
  }
  vector<int> val(n);
  for (auto &c : val) {
    c = gen() % mod;
  }
  for (int x = 1; x <= k; x++) {
    for (int y = 1; y <= x; y++) {
      for (auto &c : S[x][y]) {
        hashes[x][y] += val[c];
        hashes[x][y] %= mod;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    need += val[i];
    need %= mod;
  }
  gener(1, 0);
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
