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;
using ll = long long;
const int N = 200005;
int n, m, k;
vector<pair<int, int>> g[N];
int id[10][10], ne;
vector<int> rest[N];
ll adj[50];
void build() {
  for (int i = 1; i <= n; i++) {
    sort(g[i].begin(), g[i].end(), [](pair<int, int> a, pair<int, int> b) {
      return a.second < b.second;
    });
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      id[i][j] = ne++;
    }
  }
  for (int i = 1; i <= n; i++) {
    int sz = (int)g[i].size();
    int j = 1;
    for (auto [v, w] : g[i]) {
      rest[v].push_back(id[sz][j]);
      j++;
    }
  }
  for (int i = 0; i < ne; i++) adj[i] = (1LL << ne) - 1;
  for (int i = 1; i <= n; i++) {
    sort(rest[i].begin(), rest[i].end());
    set<int> sb;
    for (int a : rest[i]) {
      for (int b : sb) {
        adj[a] &= ~(1LL << b);
        adj[b] &= ~(1LL << a);
      }
      sb.insert(a);
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++)
      for (int w = 1; w <= i; w++)
        if (w != j) {
          adj[id[i][j]] &= ~(1LL << id[i][w]);
        }
  }
}
int ans;
void solve(int i, ll can, ll take) {
  if ((take & can) != take) return;
  if (i == k + 1) {
    ans++;
    return;
  }
  for (int j = 1; j <= i; j++) {
    solve(i + 1, can & adj[id[i][j]], take | (1LL << id[i][j]));
  }
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    int u, v, w;
    scanf("%d %d %d", &u, &v, &w);
    g[u].emplace_back(v, w);
  }
  build();
  solve(1, (1LL << ne) - 1, 0);
  cout << ans << endl;
}
