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;
using pii = pair<int, int>;
template <typename T>
using min_queue = priority_queue<T, vector<T>, greater<T>>;
int N, M, K;
vector<pii> g[200100];
vector<pii> prs[200100];
vector<int> rev[200100];
set<pii> exc[10][10];
bool mark[10][10];
vector<pii> tmp[10];
bool bad[10][10];
int res = 0;
void recur(int k) {
  if (k > K) {
    res++;
    return;
  }
  for (int j = 1; j <= k; j++) {
    if (mark[k][j] || bad[k][j]) continue;
    tmp[k].clear();
    for (auto [kk, jj] : exc[k][j]) {
      if (!mark[kk][jj]) {
        mark[kk][jj] = 1;
        tmp[k].push_back({kk, jj});
      }
    }
    recur(k + 1);
    for (auto [kk, jj] : tmp[k]) {
      mark[kk][jj] = 0;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M >> K;
  for (int i = 0; i < M; i++) {
    int a, b, w;
    cin >> a >> b >> w;
    g[a].push_back({w, b});
    rev[b].push_back(a);
  }
  for (int i = 1; i <= N; i++) sort(g[i].begin(), g[i].end());
  for (int i = 1; i <= N; i++) {
    for (int j = 0; j < g[i].size(); j++) {
      prs[g[i][j].second].push_back({g[i].size(), j + 1});
    }
  }
  set<pii> st;
  for (int v = 1; v <= N; v++) {
    st.clear();
    auto& vv = prs[v];
    for (auto [k, j] : vv) {
      if (st.count({k, j})) bad[k][j] = 1, st.erase({k, j});
      if (!bad[k][j]) st.insert({k, j});
    }
    for (auto [k, j] : st) {
      for (auto [kk, jj] : st) {
        if (k == kk && j == jj) continue;
        exc[k][j].insert({kk, jj});
      }
    }
  }
  recur(1);
  cout << res << '\n';
  return 0;
}
