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;
ifstream fin("in.in");
ofstream fout("out.out");
struct muchie {
  int node, weight, degree;
  bool operator<(const muchie &other) const { return weight < other.weight; }
  muchie(int a, int b, int c) {
    node = a;
    weight = b;
    degree = c;
  }
};
vector<muchie> out[200005], in[200005];
int q[10][10], illegal[10][10][10][10], st[10], t, n, ans, m, k, cant[10][10],
    OutDegree[200005], pre[200005][10];
void allperms(int poz) {
  if (poz > k) {
    ans++;
    return;
  }
  for (int ToNode = 1; ToNode <= poz; ToNode++) {
    if (cant[poz][ToNode]) continue;
    bool ok = 1;
    for (int j = 1; j <= poz - 1; j++) {
      if (illegal[j][st[j]][poz][ToNode]) {
        ok = 0;
        break;
      }
    }
    if (ok) {
      st[poz] = ToNode;
      allperms(poz + 1);
    }
  }
}
void Clear_q(int q[10][10]) {
  for (int i = 0; i < 10; i++)
    for (int j = 0; j < 10; j++) q[i][j] = 0;
}
int main() {
  int i, j, node;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    out[u].push_back({v, w, 0});
  }
  for (i = 1; i <= n; i++) sort(out[i].begin(), out[i].end());
  for (i = 1; i <= n; i++) {
    for (j = 0; j < out[i].size(); j++) out[i][j].degree = j + 1;
  }
  for (i = 1; i <= n; i++) {
    for (auto k : out[i]) {
      in[k.node].push_back({i, k.weight, k.degree});
    }
  }
  for (i = 1; i <= n; i++) OutDegree[i] = out[i].size();
  for (node = 1; node <= n; node++) {
    Clear_q(q);
    vector<pair<int, int>> v;
    for (i = 0; i < in[node].size(); i++) {
      int node1 = in[node][i].node;
      int a = OutDegree[node1];
      int b = in[node][i].degree;
      if (q[a][b] == 0) {
        q[a][b] = 1;
        v.push_back({a, b});
      } else
        cant[a][b] = true;
    }
    for (i = 0; i < v.size(); i++) {
      for (j = 0; j < v.size(); j++) {
        if (i == j) continue;
        illegal[v[i].first][v[i].second][v[j].first][v[j].second] = true;
      }
    }
  }
  allperms(1);
  cout << ans << "\n";
}
