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 int64_t M1 = 998244353;
const int64_t M2 = 1000000007;
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
vector<pair<int64_t, int64_t>> node[200001];
int64_t a[200001];
int64_t nxt[10][10];
int64_t sum, k, ans;
void dfs(int64_t i, int64_t x) {
  if (i == k + 1) {
    ans += (sum == x);
    return;
  }
  for (int64_t j = 0; j < i; j++) dfs(i + 1, x ^ nxt[i][j]);
}
void solve() {
  int64_t n, m;
  cin >> n >> m >> k;
  while (m--) {
    int64_t u, v, w;
    cin >> u >> v >> w;
    node[u].push_back({w, v});
  }
  sum = 0;
  for (int64_t i = 1; i <= n; i++) {
    a[i] = rng();
    sum ^= a[i];
  }
  for (int64_t u = 1; u <= n; u++) {
    if (node[u].empty()) {
      cout << 0 << "\n";
      return;
    }
    sort(node[u].begin(), node[u].end());
    for (int64_t j = 0; j < (int64_t)node[u].size(); j++)
      nxt[(int64_t)node[u].size()][j] ^= a[node[u][j].second];
  }
  ans = 0;
  dfs(1, 0);
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
}
