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>
#pragma GCC optimize("-O2")
using namespace std;
void err(istream_iterator<string> it) { cerr << endl; }
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\t";
  err(++it, args...);
}
template <typename T1, typename T2>
ostream& operator<<(ostream& c, pair<T1, T2>& v) {
  c << "(" << v.first << "," << v.second << ")";
  return c;
}
template <template <class...> class TT, class... T>
ostream& operator<<(ostream& out, TT<T...>& c) {
  out << "{ ";
  for (auto& x : c) out << x << " ";
  out << "}";
  return out;
}
const int LIM = 2e5 + 5, MOD = 1e9 + 7;
const long double EPS = 1e-9;
int read() {
  int xx = 0, ff = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    xx = (xx << 3) + (xx << 1) + ch - '0';
    ch = getchar();
  }
  return xx * ff;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long n, m, k;
vector<pair<int, int> > adj[LIM];
int deg[LIM];
vector<pair<int, int> > adj2[LIM];
bool visited[LIM];
bool po;
vector<int> check;
int curr[10];
bool pos[10][10][10][10];
int ans = 0;
void solve(int posi) {
  if (posi == k) {
    for (int i = 0; i < k; ++i) {
      for (int j = i; j < k; ++j) {
        int deg1 = i + 1, deg2 = j + 1;
        int tk1 = curr[i], tk2 = curr[j];
        if (pos[deg1][tk1][deg2][tk2] == 0) return;
      }
    }
    ans++;
    return;
  }
  for (int i = 1; i < posi + 2; ++i) {
    curr[posi] = i;
    solve(posi + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    u--;
    v--;
    deg[u]++;
    adj[u].push_back(make_pair(w, v));
  }
  for (int i = 0; i < n; ++i) {
    sort(adj[i].begin(), adj[i].end());
    for (int j = 0; j < adj[i].size(); ++j) {
      int ind = j + 1;
      int u = i, v = adj[i][j].second;
      adj2[v].push_back(make_pair(ind, deg[i]));
    }
  }
  for (int i = 0; i < 10; ++i) {
    for (int j = 0; j < 10; ++j) {
      for (int k = 0; k < 10; ++k) {
        for (int l = 0; l < 10; ++l) {
          pos[i][j][k][l] = 1;
        }
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    int pres[10][10];
    for (int j = 0; j < 10; ++j)
      for (int k = 0; k < 10; ++k) pres[j][k] = 0;
    for (int j = 0; j < adj2[i].size(); ++j) {
      int ind1 = adj2[i][j].second, sz1 = adj2[i][j].first;
      pres[ind1][sz1]++;
    }
    for (int j = 1; j < 10; ++j) {
      for (int k = 1; k < j + 1; ++k) {
        if (pres[j][k] == 0) continue;
        if (pres[j][k] >= 2) {
          pos[j][k][j][k] = 0;
        }
        for (int l = 1; l < j; ++l) {
          for (int m = 1; m < l + 1; ++m) {
            if (pres[l][m] == 0) continue;
            pos[j][k][l][m] = 0;
            pos[l][m][j][k] = 0;
          }
        }
      }
    }
  }
  vector<int> curr;
  solve(0);
  cout << ans << '\n';
  return 0;
}
