#include <bits/stdc++.h>
using namespace std;
long long n, m, k;
vector<vector<pair<long long, long long>>> graph;
long long ok[10][10];
long long fine[10][10][10][10];
bitset<(long long)(2e5 + 5)> incoming[10][10];
long long answer = 0;
vector<long long> temp;
void dfs() {
  if (temp.size() == k) {
    answer++;
    return;
  }
  long long ind = temp.size() + 1;
  for (long long i = 0; i < ind; i++) {
    bool flag = true;
    for (long long j = 1; j < ind; j++) {
      if (!fine[ind][i][j][temp[j - 1]]) {
        flag = false;
        break;
      }
    }
    if (!flag) continue;
    if (ok[ind][i]) {
      temp.push_back(i);
      dfs();
      temp.pop_back();
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m >> k;
  graph.resize(n + 1);
  for (long long i = 0; i < m; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    graph[u].push_back({w, v});
  }
  for (long long i = 1; i < n + 1; i++) {
    if (graph[i].size() == 0) {
      cout << 0 << '\n';
      return 0;
    }
    sort(graph[i].begin(), graph[i].end());
  }
  for (long long i = 1; i < k + 1; i++)
    for (long long j = 0; j < i; j++) ok[i][j] = 1;
  for (long long out = 1; out < k + 1; out++) {
    for (long long i = 1; i < n + 1; i++) {
      if (graph[i].size() == out) {
        for (long long j = 0; j < graph[i].size(); j++) {
          long long ver = graph[i][j].second;
          if (incoming[out][j][ver]) {
            ok[out][j] = 0;
            continue;
          }
          incoming[out][j][ver] = 1;
        }
      }
    }
  }
  for (long long a = 1; a < k + 1; a++)
    for (long long b = 0; b < a; b++)
      for (long long c = a; c < k + 1; c++)
        for (long long d = 0; d < c; d++) {
          if ((incoming[a][b] & incoming[c][d]) == 0) {
            fine[a][b][c][d] = fine[c][d][a][b] = 1;
          }
        }
  dfs();
  cout << answer << '\n';
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, m, k;
vector<vector<pair<long long, long long>>> graph;
long long ok[10][10];
long long fine[10][10][10][10];
bitset<(long long)(2e5 + 5)> incoming[10][10];
long long answer = 0;
vector<long long> temp;
void dfs() {
  if (temp.size() == k) {
    answer++;
    return;
  }
  long long ind = temp.size() + 1;
  for (long long i = 0; i < ind; i++) {
    bool flag = true;
    for (long long j = 1; j < ind; j++) {
      if (!fine[ind][i][j][temp[j - 1]]) {
        flag = false;
        break;
      }
    }
    if (!flag) continue;
    if (ok[ind][i]) {
      temp.push_back(i);
      dfs();
      temp.pop_back();
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m >> k;
  graph.resize(n + 1);
  for (long long i = 0; i < m; i++) {
    long long u, v, w;
    cin >> u >> v >> w;
    graph[u].push_back({w, v});
  }
  for (long long i = 1; i < n + 1; i++) {
    if (graph[i].size() == 0) {
      cout << 0 << '\n';
      return 0;
    }
    sort(graph[i].begin(), graph[i].end());
  }
  for (long long i = 1; i < k + 1; i++)
    for (long long j = 0; j < i; j++) ok[i][j] = 1;
  for (long long out = 1; out < k + 1; out++) {
    for (long long i = 1; i < n + 1; i++) {
      if (graph[i].size() == out) {
        for (long long j = 0; j < graph[i].size(); j++) {
          long long ver = graph[i][j].second;
          if (incoming[out][j][ver]) {
            ok[out][j] = 0;
            continue;
          }
          incoming[out][j][ver] = 1;
        }
      }
    }
  }
  for (long long a = 1; a < k + 1; a++)
    for (long long b = 0; b < a; b++)
      for (long long c = a; c < k + 1; c++)
        for (long long d = 0; d < c; d++) {
          if ((incoming[a][b] & incoming[c][d]) == 0) {
            fine[a][b][c][d] = fine[c][d][a][b] = 1;
          }
        }
  dfs();
  cout << answer << '\n';
}

```