#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int>> adj[200005];
int bad[9][9][9][9];
int cnt[200005][9][9];
int worst[9][9];
int n, m, k;
int get(int x, vector<int> a) {
  if (x == k) {
    return 1;
  }
  int res = 0;
  for (int i = 0; i <= x; ++i) {
    if (worst[x][i]) continue;
    bool poss = true;
    for (int j = 0; j < x && poss; ++j) {
      if (bad[x][i][j][a[j]] > 0) {
        poss = false;
      }
    }
    if (poss) {
      vector<int> b = a;
      b.push_back(i);
      res += get(x + 1, b);
    }
  }
  return res;
}
void solve(int tc) {
  cin >> n >> m >> k;
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    adj[u - 1].push_back({w - 1, v - 1});
  }
  for (int i = 0; i < n; ++i) {
    sort((adj[i]).begin(), (adj[i]).end());
    int outdeg = adj[i].size();
    for (int j = 0; j < outdeg; ++j) {
      int v = adj[i][j].second;
      cnt[v][outdeg - 1][j]++;
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j1 = 0; j1 < 9; ++j1) {
      for (int k1 = 0; k1 < 9; ++k1) {
        if (cnt[i][j1][k1] == 0) continue;
        if (cnt[i][j1][k1] > 1) {
          worst[j1][k1] = 1;
          continue;
        }
        for (int j2 = 0; j2 < 9; ++j2) {
          for (int k2 = 0; k2 < 9; ++k2) {
            if (j1 == j2 && k1 == k2) continue;
            if (cnt[i][j2][k2] > 0) {
              bad[j1][k1][j2][k2] = 1;
              bad[j2][k2][j1][k1] = 1;
            }
          }
        }
      }
    }
  }
  cout << get(0, vector<int>()) << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  for (int tc = 1; tc <= t; ++tc) {
    solve(tc);
  }
  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>
using namespace std;
vector<pair<int, int>> adj[200005];
int bad[9][9][9][9];
int cnt[200005][9][9];
int worst[9][9];
int n, m, k;
int get(int x, vector<int> a) {
  if (x == k) {
    return 1;
  }
  int res = 0;
  for (int i = 0; i <= x; ++i) {
    if (worst[x][i]) continue;
    bool poss = true;
    for (int j = 0; j < x && poss; ++j) {
      if (bad[x][i][j][a[j]] > 0) {
        poss = false;
      }
    }
    if (poss) {
      vector<int> b = a;
      b.push_back(i);
      res += get(x + 1, b);
    }
  }
  return res;
}
void solve(int tc) {
  cin >> n >> m >> k;
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    adj[u - 1].push_back({w - 1, v - 1});
  }
  for (int i = 0; i < n; ++i) {
    sort((adj[i]).begin(), (adj[i]).end());
    int outdeg = adj[i].size();
    for (int j = 0; j < outdeg; ++j) {
      int v = adj[i][j].second;
      cnt[v][outdeg - 1][j]++;
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j1 = 0; j1 < 9; ++j1) {
      for (int k1 = 0; k1 < 9; ++k1) {
        if (cnt[i][j1][k1] == 0) continue;
        if (cnt[i][j1][k1] > 1) {
          worst[j1][k1] = 1;
          continue;
        }
        for (int j2 = 0; j2 < 9; ++j2) {
          for (int k2 = 0; k2 < 9; ++k2) {
            if (j1 == j2 && k1 == k2) continue;
            if (cnt[i][j2][k2] > 0) {
              bad[j1][k1][j2][k2] = 1;
              bad[j2][k2][j1][k1] = 1;
            }
          }
        }
      }
    }
  }
  cout << get(0, vector<int>()) << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  for (int tc = 1; tc <= t; ++tc) {
    solve(tc);
  }
  return 0;
}

```