#include <bits/stdc++.h>
using namespace std;
int n, m, k, ch[10][10][10][10], per[10];
vector<pair<int, int>> out[200001];
int in[200001][10][10];
int solve(int cur) {
  if (cur == k) {
    for (int i = 1; i <= k; i++) {
      for (int j = i; j <= k; j++) {
        if (ch[i][per[i]][j][per[j]]) {
          return 0;
        }
      }
    }
    return 1;
  }
  int ret = 0;
  for (int i = 1; i <= cur + 1; i++) {
    per[cur + 1] = i;
    ret += solve(cur + 1);
    per[cur + 1] = 0;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    out[u].push_back({w, v});
  }
  for (int i = 1; i <= n; i++) {
    sort(out[i].begin(), out[i].end());
    int sz = out[i].size();
    for (int j = 1; j <= sz; j++) {
      in[out[i][j - 1].second][sz][j]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int a = 1; a <= k; a++) {
      for (int b = 1; b <= a; b++) {
        for (int c = 1; c <= k; c++) {
          for (int d = 1; d <= c; d++) {
            if (a == c && b == d) continue;
            if (in[i][a][b] != 0 && in[i][c][d] != 0) ch[a][b][c][d] = 1;
          }
        }
        if (in[i][a][b] >= 2) ch[a][b][a][b] = 1;
      }
    }
  }
  cout << solve(0);
  return 0;
}

### Prompt

Please create a solution in cpp to 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;
int n, m, k, ch[10][10][10][10], per[10];
vector<pair<int, int>> out[200001];
int in[200001][10][10];
int solve(int cur) {
  if (cur == k) {
    for (int i = 1; i <= k; i++) {
      for (int j = i; j <= k; j++) {
        if (ch[i][per[i]][j][per[j]]) {
          return 0;
        }
      }
    }
    return 1;
  }
  int ret = 0;
  for (int i = 1; i <= cur + 1; i++) {
    per[cur + 1] = i;
    ret += solve(cur + 1);
    per[cur + 1] = 0;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    out[u].push_back({w, v});
  }
  for (int i = 1; i <= n; i++) {
    sort(out[i].begin(), out[i].end());
    int sz = out[i].size();
    for (int j = 1; j <= sz; j++) {
      in[out[i][j - 1].second][sz][j]++;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int a = 1; a <= k; a++) {
      for (int b = 1; b <= a; b++) {
        for (int c = 1; c <= k; c++) {
          for (int d = 1; d <= c; d++) {
            if (a == c && b == d) continue;
            if (in[i][a][b] != 0 && in[i][c][d] != 0) ch[a][b][c][d] = 1;
          }
        }
        if (in[i][a][b] >= 2) ch[a][b][a][b] = 1;
      }
    }
  }
  cout << solve(0);
  return 0;
}

```