#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
template <typename T1, typename T2>
inline void chmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void chmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
using namespace std;
signed main() {
  std::cin.tie(nullptr);
  std::ios_base::sync_with_stdio(false);
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<pair<int, int>>> G(n);
  vector<vector<pair<int, int>>> invG(n);
  for (int i = 0; i < (m); i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    G[a].push_back({c, b});
    invG[b].push_back({c, a});
  }
  for (int i = 0; i < (n); i++) sort((G[i]).begin(), (G[i]).end());
  vector<vector<int>> v(k + 1);
  for (int i = 0; i < (n); i++) {
    v[G[i].size()].push_back(i);
    if (invG[i].size() == 0 || G[i].size() == 0) {
      cout << 0 << endl;
      return 0;
    }
  }
  bool ok[k + 1][k + 1][k + 1][k + 1];
  for (int i = 0; i < (k + 1); i++)
    for (int j = 0; j < (k + 1); j++)
      for (int x = 0; x < (k + 1); x++)
        for (int y = 0; y < (k + 1); y++) ok[i][j][x][y] = true;
  for (int i = 1; i <= k; i++) {
    for (int j = i + 1; j <= k; j++) {
      for (int x = 1; x <= i; x++) {
        for (int y = 1; y <= j; y++) {
          vector<bool> use(n, 0);
          for (auto u : v[i]) {
            if (use[G[u][x - 1].second]) ok[i][j][x][y] = false;
            use[G[u][x - 1].second] = 1;
          }
          for (auto u : v[j]) {
            if (use[G[u][y - 1].second]) ok[i][j][x][y] = false;
            use[G[u][y - 1].second] = 1;
          }
        }
      }
    }
  }
  int ans = 0;
  vector<vector<int>> vec;
  vector<int> st;
  st.push_back(1);
  vec.push_back(st);
  while (true) {
    int sz = vec[0].size();
    if (sz == k) break;
    vector<vector<int>> res;
    for (auto u : vec) {
      for (int i = 1; i <= sz + 1; i++) {
        u.push_back(i);
        res.push_back(u);
        u.pop_back();
      }
    }
    vec = res;
  }
  for (auto u : vec) {
    int flg = 1;
    for (int i = 1; i <= k; i++) {
      for (int j = 1; j <= k; j++) {
        if (!ok[i][j][u[i - 1]][u[j - 1]]) flg = 0;
      }
    }
    ans += flg;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
template <typename T1, typename T2>
inline void chmin(T1 &a, T2 b) {
  if (a > b) a = b;
}
template <typename T1, typename T2>
inline void chmax(T1 &a, T2 b) {
  if (a < b) a = b;
}
using namespace std;
signed main() {
  std::cin.tie(nullptr);
  std::ios_base::sync_with_stdio(false);
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<pair<int, int>>> G(n);
  vector<vector<pair<int, int>>> invG(n);
  for (int i = 0; i < (m); i++) {
    int a, b, c;
    cin >> a >> b >> c;
    a--;
    b--;
    G[a].push_back({c, b});
    invG[b].push_back({c, a});
  }
  for (int i = 0; i < (n); i++) sort((G[i]).begin(), (G[i]).end());
  vector<vector<int>> v(k + 1);
  for (int i = 0; i < (n); i++) {
    v[G[i].size()].push_back(i);
    if (invG[i].size() == 0 || G[i].size() == 0) {
      cout << 0 << endl;
      return 0;
    }
  }
  bool ok[k + 1][k + 1][k + 1][k + 1];
  for (int i = 0; i < (k + 1); i++)
    for (int j = 0; j < (k + 1); j++)
      for (int x = 0; x < (k + 1); x++)
        for (int y = 0; y < (k + 1); y++) ok[i][j][x][y] = true;
  for (int i = 1; i <= k; i++) {
    for (int j = i + 1; j <= k; j++) {
      for (int x = 1; x <= i; x++) {
        for (int y = 1; y <= j; y++) {
          vector<bool> use(n, 0);
          for (auto u : v[i]) {
            if (use[G[u][x - 1].second]) ok[i][j][x][y] = false;
            use[G[u][x - 1].second] = 1;
          }
          for (auto u : v[j]) {
            if (use[G[u][y - 1].second]) ok[i][j][x][y] = false;
            use[G[u][y - 1].second] = 1;
          }
        }
      }
    }
  }
  int ans = 0;
  vector<vector<int>> vec;
  vector<int> st;
  st.push_back(1);
  vec.push_back(st);
  while (true) {
    int sz = vec[0].size();
    if (sz == k) break;
    vector<vector<int>> res;
    for (auto u : vec) {
      for (int i = 1; i <= sz + 1; i++) {
        u.push_back(i);
        res.push_back(u);
        u.pop_back();
      }
    }
    vec = res;
  }
  for (auto u : vec) {
    int flg = 1;
    for (int i = 1; i <= k; i++) {
      for (int j = 1; j <= k; j++) {
        if (!ok[i][j][u[i - 1]][u[j - 1]]) flg = 0;
      }
    }
    ans += flg;
  }
  cout << ans << endl;
  return 0;
}

```