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 int maxn = 2e5 + 5;
const int maxk = 10;
const int mod = 19260817;
int n, m, k, u, v, w, ans;
priority_queue<pair<int, int>, vector<pair<int, int> >,
               greater<pair<int, int> > >
    pq[maxn];
long long hsh[maxn], all_hash, single_hash[maxk][maxk];
void dfs(int pos, long long now) {
  if (pos > k) {
    if (now == all_hash) ++ans;
    return;
  }
  for (int i = 1; i <= pos; ++i) {
    dfs(pos + 1, now + single_hash[pos][i]);
  }
}
int main() {
  cin >> n >> m >> k;
  while (m--) {
    cin >> u >> v >> w;
    pq[u].push(pair<int, int>(w, v));
  }
  srand(time(NULL));
  for (int i = 1; i <= n; ++i) {
    hsh[i] = rand() % mod;
    all_hash += hsh[i];
  }
  for (int i = 1; i <= n; ++i) {
    int deg = pq[i].size();
    for (int j = 1; j <= deg; ++j) {
      single_hash[deg][j] += hsh[pq[i].top().second];
      pq[i].pop();
    }
  }
  dfs(1, 0);
  cout << ans;
  return 0;
}
