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>
const int N = 2e5 + 5;
using namespace std;
int n, m, k, c[10], res;
int kt[10][10][10][10];
vector<int> b[10];
set<int> ver[10][10];
vector<pair<int, int> > a[N];
void Try(int d) {
  if (d == k + 1) {
    int check = 1;
    for (int i = 1; i <= k; i++)
      for (int j = i + 1; j <= k; j++)
        if (!kt[i][c[i]][j][c[j]]) i = k + 1, j = k + 1, check = 0;
    for (int i = 1; i <= k; i++)
      if (*ver[i][c[i]].begin() == -1) check = 0, i = k + 1;
    if (check) res++;
    return;
  }
  for (int i = 1; i <= d; i++) {
    c[d] = i;
    Try(d + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    a[u].push_back(make_pair(w, v));
  }
  for (int i = 1; i <= n; i++) {
    b[a[i].size()].push_back(i);
    sort(a[i].begin(), a[i].end());
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      for (int u : b[i]) {
        if (ver[i][j].find(a[u][j - 1].second) != ver[i][j].end()) {
          ver[i][j].clear();
          ver[i][j].insert(-1);
          break;
        }
        ver[i][j].insert(a[u][j - 1].second);
      }
    }
  }
  for (int i1 = 1; i1 <= k; i1++) {
    for (int j1 = 1; j1 <= i1; j1++) {
      if (*ver[i1][j1].begin() == -1) continue;
      for (int i2 = i1 + 1; i2 <= k; i2++) {
        for (int j2 = 1; j2 <= i2; j2++) {
          if (*ver[i2][j2].begin() == -1) continue;
          for (int u : ver[i2][j2]) {
            if (ver[i1][j1].find(u) != ver[i1][j1].end()) {
              kt[i1][j1][i2][j2] = 1;
              break;
            }
          }
          kt[i1][j1][i2][j2] = 1 - kt[i1][j1][i2][j2];
        }
      }
    }
  }
  Try(1);
  cout << res;
}
