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 mod = 1000000007;
const vector<int> mods = {998244353, 1000000007, 1000000009};
const long long inf = 3e18;
const double pi = acos(-1.0);
const double eps = 0.0000001;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <class... K>
using umap = unordered_map<K...>;
template <class... K>
using uset = unordered_set<K...>;
int n, m, k;
int u, v, w;
const int N = 200100;
const int P = 400000;
umap<int, vector<pair<int, int> > > g;
long long pw[N][3];
long long fc[10][10][3];
vector<long long> goal;
long long comb[15];
long long res;
void build() {
  for (int j = (0); j < (3); ++j) pw[0][j] = 1;
  for (int i = (1); i < (n + 10); ++i)
    for (int j = (0); j < (3); ++j) {
      pw[i][j] = (pw[i - 1][j] * 13) % mods[j];
    }
  goal.resize(3);
  for (int i = (0); i < (n); ++i)
    for (int j = (0); j < (3); ++j) {
      goal[j] = (goal[j] + pw[i][j]) % mods[j];
    }
}
void inc() {
  long long car = 1;
  for (int i = (1); i < (10); ++i) {
    long long nxc = (comb[i] + car) / i;
    comb[i] = (comb[i] + car) % i;
    car = nxc;
  }
}
bool alz() {
  for (int i = (1); i < (k + 1); ++i) {
    if (comb[i]) return 0;
  }
  return 1;
}
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.precision(13);
  cin >> n >> m >> k;
  build();
  for (int i = (0); i < (m); ++i) {
    cin >> u >> v >> w;
    u--;
    v--;
    g[u].push_back({w, v});
  }
  for (int i = (0); i < (n); ++i) sort(begin(g[i]), end(g[i]));
  for (int i = (0); i < (n); ++i) {
    long long sg = ((int)(g[i]).size());
    for (int j = (0); j < (sg); ++j)
      for (int k = (0); k < (3); ++k) {
        fc[sg][j][k] = (fc[sg][j][k] + pw[g[i][j].second][k]) % mods[k];
      }
  }
  res = 0;
  for (int i = (0); i < (P); ++i) {
    vector<long long> cur(3);
    for (int i = (0); i < (k); ++i)
      for (int j = (0); j < (3); ++j) {
        cur[j] = (cur[j] + fc[i + 1][comb[i + 1]][j]) % mods[j];
      }
    if (cur == goal) res++;
    inc();
    if (alz()) break;
  }
  cout << res << "\n";
  return 0;
}
