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 MN = 2e5 + 10;
const int MK = 12;
int N, M, K, f;
struct Edge {
 public:
  int n, w;
  bool operator<(const Edge& o) const { return w < o.w; }
};
struct Ord {
 public:
  int i, k;
  bool operator<(const Ord& o) const {
    return k < o.k || !(o.k < k) && i < o.i;
  }
  bool operator==(const Ord& o) const { return i == o.i && k == o.k; }
};
std::vector<Edge> a[MN];
std::vector<Ord> b[MN];
std::vector<Ord> rules[MK][MK];
int ok[MK][MK];
void brute(int k = 1) {
  if (k == K + 1) return (void)(++f);
  for (int i = 0; i < k; ++i)
    if (ok[k][i] == 0) {
      for (auto x : rules[k][i]) ++ok[x.k][x.i];
      brute(k + 1);
      for (auto x : rules[k][i]) --ok[x.k][x.i];
    }
}
int main() {
  scanf("%d%d%d", &N, &M, &K);
  for (int i = 0, u, v, w; i < M; ++i)
    scanf("%d%d%d", &u, &v, &w), a[u].push_back({v, w});
  for (int i = 1; i <= N; ++i) std::sort(a[i].begin(), a[i].end());
  for (int i = 1; i <= N; ++i)
    for (int j = 0; j < a[i].size(); ++j)
      b[a[i][j].n].push_back({j, (int)a[i].size()});
  for (int i = 1; i <= N; ++i) {
    std::sort(b[i].begin(), b[i].end());
    for (int j = 0; j + 1 < b[i].size(); ++j)
      if (b[i][j] == b[i][j + 1]) ok[b[i][j].k][b[i][j].i] = 1;
    b[i].resize(
        std::distance(b[i].begin(), std::unique(b[i].begin(), b[i].end())));
    for (int j = 0; j < b[i].size(); ++j)
      for (int k = j + 1; k < b[i].size(); ++k)
        if (b[i][j].k < b[i][k].k) {
          rules[b[i][j].k][b[i][j].i].push_back(b[i][k]);
        }
  }
  for (int i = 1; i <= K; ++i)
    for (int j = 0; j < i; ++j) {
      std::sort(rules[i][j].begin(), rules[i][j].end());
      rules[i][j].resize(
          std::distance(rules[i][j].begin(),
                        std::unique(rules[i][j].begin(), rules[i][j].end())));
    }
  brute();
  printf("%d\n", f);
  return 0;
}
