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;
mt19937 rnd((size_t)time(nullptr));
const long long B[] = {43, 101, 433};
const long long M[] = {1000000007, 1000013029, 1000000009};
long long st(long long a, long long b, long long M) {
  if (b == 0) {
    return 1;
  }
  if (b % 2) {
    return (a * st((a * a) % M, b / 2, M)) % M;
  } else {
    return st((a * a) % M, b / 2, M);
  }
}
struct hash_num {
  long long res[3];
  hash_num() {
    for (int i = 0; i < 3; ++i) {
      res[i] = 0;
    }
  }
  void operator+=(hash_num &h) {
    for (int i = 0; i < 3; ++i) {
      res[i] += h.res[i];
      res[i] %= M[i];
    }
  }
  void add_num(int num) {
    for (int i = 0; i < 3; ++i) {
      res[i] += st(B[i], num, M[i]);
      res[i] %= M[i];
    }
  }
  bool operator==(hash_num &h) {
    for (int i = 0; i < 3; ++i) {
      if (res[i] != h.res[i]) {
        return false;
      }
    }
    return true;
  }
};
const int MAXN = 200100, MAXK = 12;
int n, m, k;
vector<pair<int, int> > g[MAXN];
set<int> m_w[MAXK][MAXK];
hash_num h_s[MAXK][MAXK];
hash_num need_h;
long long ans = 0;
hash_num calc_h(set<int> &w) {
  hash_num res;
  for (int i : w) {
    res.add_num(i);
  }
  return res;
}
void dfs(int cur_i, hash_num cur_h) {
  if (cur_i == k + 1) {
    if (cur_h == need_h) ++ans;
    return;
  }
  for (int j = 0; j < cur_i; ++j) {
    hash_num n_h = cur_h;
    n_h += h_s[cur_i][j];
    dfs(cur_i + 1, n_h);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    u--;
    v--;
    g[u].push_back({w, v});
  }
  for (int i = 0; i < n; ++i) {
    sort(g[i].begin(), g[i].end());
    for (int j = 0; j < g[i].size(); ++j) {
      m_w[g[i].size()][j].insert(g[i][j].second);
    }
  }
  for (int i = 0; i < n; ++i) {
    need_h.add_num(i);
  }
  for (int i = 1; i <= k; ++i) {
    for (int j = 0; j < i; ++j) {
      h_s[i][j] = calc_h(m_w[i][j]);
    }
  }
  dfs(1, hash_num());
  cout << ans;
  return 0;
}
