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 M = 1e9 + 7;
const long long mod = 998244353;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long FIXED_RANDOM =
    chrono::steady_clock::now().time_since_epoch().count();
struct Hash {
  long long val1 = 1;
  Hash() {}
  Hash(long long a) { val1 = a; }
  Hash operator*(const Hash &nxt) const { return Hash(val1 * nxt.val1 % M); }
  void operator=(const Hash &nxt) { val1 = nxt.val1; }
  void operator*=(const Hash &nxt) { val1 = nxt.val1 * val1 % M; }
  Hash calc_Hash(const long long &x) { return (x + FIXED_RANDOM) % M; }
  bool operator==(const Hash &nxt) const { return val1 == nxt.val1; }
};
int n, m, k;
vector<pair<int, int> > adj[200200];
Hash vals[200200], tot, S[10][10];
int out[200200];
int ans = 0;
void recurse(int i, Hash h) {
  if (i == k + 1) {
    if (h == tot) ans++;
    return;
  }
  for (int j = 1; j <= i; j++) {
    recurse(i + 1, S[i][j] * h);
  }
}
void solve() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) {
    vals[i] = vals[i].calc_Hash(i);
    tot *= vals[i];
  }
  for (int i = 0; i < m; i++) {
    int a, b, w;
    cin >> a >> b >> w;
    adj[a].push_back({b, w});
    out[a]++;
  }
  for (int i = 1; i <= n; i++) {
    sort(adj[i].begin(), adj[i].end(), [&](pair<int, int> a, pair<int, int> b) {
      return a.second < b.second;
    });
    for (int j = 0; j < out[i]; j++) {
      S[out[i]][j + 1] *= vals[adj[i][j].first];
    }
  }
  Hash h;
  recurse(1, h);
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}
