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 double PI = 2 * acos(0.0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());
const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";
template <class T0, class T1>
inline ostream &operator<<(ostream &out, pair<T0, T1> &a) {
  return out << "{" << a.first << ", " << a.second << "}";
}
template <class T0, class T1>
inline istream &operator>>(istream &in, pair<T0, T1> &a) {
  return in >> a.first >> a.second;
}
template <class T0, class T1, class T2>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << "}";
}
template <class T0, class T1, class T2, class T3>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2, T3> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << ", " << get<3>(a) << "}";
}
template <class T>
inline ostream &operator<<(ostream &out, vector<T> &a) {
  out << "[";
  for (int i = 0; i < int(a.size()); ++i)
    out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
  return out;
}
void smain();
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(12) << fixed;
  smain();
  return 0;
}
const int T = 10;
const int N = 2e5 + 10;
bool can[T][T][T][T];
bool forbid[T][T];
bitset<N> ptr[T][T];
struct Edge {
  int to, w;
};
vector<Edge> e;
vector<int> g[N];
long long ans = 0;
int n, m, k;
int c[T];
void rec(int i) {
  if (i == k + 1) {
    ans++;
    return;
  }
  for (int val = 1; val <= i; ++val) {
    if (forbid[i][val]) continue;
    bool ican = true;
    for (int j = 1; j < i; ++j) {
      if (!can[j][c[j]][i][val]) {
        ican = false;
      }
    }
    if (ican) {
      c[i] = val;
      rec(i + 1);
    }
  }
}
void smain() {
  cin >> n >> m >> k;
  for (int _ = 0; _ < int(m); ++_) {
    int u, v, w;
    cin >> u >> v >> w;
    g[u].push_back(e.size());
    e.push_back({v, w});
  }
  for (int v = 1; v <= n; ++v) {
    sort(g[v].begin(), g[v].end(), [&](const int &id1, const int &id2) -> bool {
      return e[id1].w < e[id2].w;
    });
  }
  for (int i = 1; i <= k; ++i) {
    for (int val = 1; val <= i; ++val) {
      for (int v = 1; v <= n; ++v) {
        if (g[v].size() != i) continue;
        int cur = e[g[v][val - 1]].to;
        if (ptr[i][val][cur]) forbid[i][val] = true;
        ptr[i][val][cur] = 1;
      }
    }
  }
  for (int i = 1; i <= k; ++i) {
    for (int vali = 1; vali <= i; ++vali) {
      for (int j = 1; j <= k; ++j) {
        for (int valj = 1; valj <= j; ++valj) {
          if ((ptr[i][vali] & ptr[j][valj]).count() == 0) {
            can[i][vali][j][valj] = true;
          }
        }
      }
    }
  }
  rec(1);
  cout << ans;
}
