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;
long long n, x, k, z, m, pos, q, ans_v, y, root, xx;
long long ii;
long long a[1000500], b[1005000], dp[1000500], used[1005000];
vector<pair<long long, long long> > g[200500];
long long bad[10][10][10][10];
set<pair<long long, long long> > ff[200500];
long long zalupa[10][10];
vector<long long> pshe;
long long ANS = 0;
void dfs(long long x) {
  if (x == k + 1) {
    bool flag = true;
    for (int i = 0; i < pshe.size(); i++) {
      for (int j = 0; j < pshe.size(); j++) {
        if (zalupa[i + 1][pshe[i]] || zalupa[j + 1][pshe[j]] ||
            bad[i + 1][pshe[i]][j + 1][pshe[j]]) {
          flag = false;
        }
      }
    }
    if (flag) {
      ANS++;
    }
    return;
  }
  for (int j = 1; j <= x; j++) {
    pshe.push_back(j);
    dfs(x + 1);
    pshe.pop_back();
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    cin >> x >> y >> z;
    g[x].push_back(make_pair(z, y));
  }
  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++) {
      auto x = g[i][j];
      long long hj = (long long)g[i].size();
      long long hk = j + 1;
      auto p = make_pair(hj, hk);
      if (ff[x.second].find(p) != ff[x.second].end()) {
        zalupa[hj][hk] = 1;
      } else {
        ff[x.second].insert(p);
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    for (auto x : ff[i]) {
      for (auto y : ff[i]) {
        if (x == y) {
          continue;
        }
        if (!bad[x.first][x.second][y.first][y.second])
          bad[x.first][x.second][y.first][y.second] = 1;
      }
    }
  }
  dfs(1);
  cout << ANS << "\n";
  return 0;
}
