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;
int n, m, k, first, second, z, ans, a[10], b[10][10][10][10];
vector<pair<int, int> > t[10][10], temp[200005];
vector<pair<int, int> > g[200005];
vector<int> h[200005];
void Gen(int p) {
  if (p == k + 1) {
    for (int i = 1; i <= k; i++) {
      for (int j = i; j <= k; j++) {
        if (b[i][a[i]][j][a[j]]) return;
      }
    }
    ans++;
    return;
  }
  for (int i = 1; i <= p; i++) {
    a[p] = i;
    Gen(p + 1);
  }
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> first >> second >> z;
    g[first].push_back({z, second});
    h[second].push_back(first);
  }
  for (int i = 1; i <= n; i++) sort((g[i]).begin(), (g[i]).end());
  for (int i = 1; i <= n; i++) {
    for (int k = 0; k < g[i].size(); k++) {
      temp[g[i][k].second].push_back({g[i].size(), k + 1});
    }
  }
  for (int i = 1; i <= n; i++) {
    for (pair<int, int> j : temp[i]) b[j.first][j.second][j.first][j.second]++;
    temp[i].erase(unique((temp[i]).begin(), (temp[i]).end()), temp[i].end());
    for (pair<int, int> j : temp[i])
      b[j.first][j.second][j.first][j.second] -= 2;
    for (pair<int, int> j : temp[i]) {
      for (pair<int, int> k : temp[i]) {
        b[j.first][j.second][k.first][k.second]++;
      }
    }
  }
  Gen(1);
  cout << ans;
}
