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;
template <typename T, typename U>
bool ckmin(T &a, U b) {
  return b < a ? a = b, true : false;
}
template <typename T, typename U>
bool ckmax(T &a, U b) {
  return b > a ? a = b, true : false;
}
unsigned seed = std::chrono::system_clock::now().time_since_epoch().count();
mt19937 gen(seed);
long long to_get = 0, res = 0;
int n, m, k;
vector<long long> hsh;
vector<vector<int>> adj;
vector<vector<long long>> out;
vector<vector<pair<int, int>>> ord;
map<pair<int, int>, int> mx;
void dfs(long long pos, long long x) {
  if (pos == k) {
    res += !(to_get ^ x);
    return;
  }
  for (int i = 1; i <= pos + 1; ++i) {
    dfs(pos + 1, x ^ out[pos + 1][i]);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m >> k;
  out.assign(k + 1, vector<long long>(k + 1, 0));
  adj.assign(n, vector<int>());
  ord.assign(n, vector<pair<int, int>>());
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    u--, v--;
    adj[u].push_back(v);
    ord[u].push_back({w, v});
  }
  uniform_int_distribution<long long> dist(0, (1LL << 63));
  for (int i = 0; i < n; ++i) {
    hsh.push_back(dist(gen));
  }
  for (auto i : hsh) to_get ^= i;
  for (int i = 0; i < n; ++i) {
    sort(ord[i].begin(), ord[i].end());
    for (int j = 0; j < (int)(ord[i]).size(); ++j) {
      mx[make_pair(i, ord[i][j].second)] = j + 1;
    }
  }
  for (auto i : mx) {
    out[(int)(adj[i.first.first]).size()][i.second] ^= hsh[i.first.second];
  }
  dfs(0, 0);
  cout << res;
  return 0;
}
