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;
struct nod {
  int to, len;
  nod() {}
  nod(int a1, int a2) { to = a1, len = a2; }
};
vector<nod> edge[200005];
bool cmp(nod i, nod j) { return i.len < j.len; }
int a[200005][105], vis[105][105], tmp[10], ans;
void dfs(int p) {
  if (p == k + 1) {
    for (int i = 1; i <= k; i++)
      for (int j = i; j <= k; j++)
        if (vis[i * 10 + tmp[i]][j * 10 + tmp[j]] == 1 ||
            vis[j * 10 + tmp[j]][i * 10 + tmp[i]] == 1)
          return;
    ans++;
    return;
  }
  for (int i = 1; i <= p; i++) tmp[p] = i, dfs(p + 1);
}
signed main() {
  ios::sync_with_stdio(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    edge[u].push_back(nod(v, w));
  }
  for (int i = 1; i <= n; i++) sort(edge[i].begin(), edge[i].end(), cmp);
  for (int i = 1; i <= n; i++) {
    int siz = edge[i].size();
    for (int j = 1; j <= siz; j++) {
      nod tt = edge[i][j - 1];
      for (int k = 1; k <= 100; k++) {
        if (a[tt.to][k]) {
          vis[k][siz * 10 + j] = vis[siz * 10 + j][k] = 1;
        }
      }
      a[tt.to][siz * 10 + j] = 1;
    }
  }
  dfs(1);
  cout << ans << endl;
}
