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>
constexpr int maxn = 200000;
constexpr int maxm = 200000;
constexpr int maxk = 9;
constexpr int ix(int i, int j) { return (i * (i - 1) >> 1) + j; }
constexpr int maxix = ix(maxk + 1, 0);
struct Edge {
  int u;
  int v;
  int w;
};
bool operator<(Edge e0, Edge e1) { return e0.w < e1.w; }
int outDegree[maxn + 1];
int usedOutDegree[maxn + 1];
Edge es[maxm];
int incl[maxn + 1][maxix];
bool taboo[maxix][maxix];
int main() {
  using namespace std;
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    es[i] = Edge{u, v, w};
    ++outDegree[u];
  }
  sort(es, es + m);
  for (int i = 0; i < m; ++i) {
    const Edge e = es[i];
    const int p = outDegree[e.u];
    int& q = usedOutDegree[e.u];
    incl[e.v][ix(p, q++)] += 1;
  }
  for (int i = 1; i <= n; ++i)
    for (int p = 0; p < maxix; ++p)
      if (incl[i][p] == 1)
        for (int q = p + 1; q < maxix; ++q)
          taboo[p][q] = taboo[p][q] || incl[i][q];
      else
        taboo[p][p] = taboo[p][p] || incl[i][p] > 1;
  int tuple[maxk + 1];
  int offsets[maxk + 1];
  for (int i = 1; i <= k; ++i) {
    tuple[i] = 0;
    offsets[i] = ix(i, 0);
  }
  auto incr = [&]() {
    for (int i = 1; i <= k; ++i)
      if (tuple[i] < i - 1) {
        ++tuple[i];
        return true;
      } else
        tuple[i] = 0;
    return false;
  };
  auto isValid = [&]() {
    for (int i = 1; i <= k; ++i) {
      const int p = tuple[i] + offsets[i];
      for (int j = i; j <= k; ++j)
        if (taboo[p][tuple[j] + offsets[j]]) return false;
    }
    return true;
  };
  int ct = 0;
  do ct += isValid();
  while (incr());
  cout << ct << '\n';
  return 0;
}
