#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
const int K = 10;
bool can[K][K][K][K], good[K][K];
int in[N][K][K];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> from(m), to(m);
  for (int i = 0; i < m; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    x--;
    y--;
    z--;
    from[z] = x;
    to[z] = y;
  }
  vector<vector<int>> out(n);
  for (int i = 0; i < m; i++) out[from[i]].push_back(to[i]);
  for (int i = 0; i < n; i++) {
    int sz = out[i].size();
    for (int j = 0; j < sz; j++) in[out[i][j]][sz][j]++;
  }
  for (int i = 1; i <= k; i++)
    for (int ci = 0; ci < i; ci++)
      for (int j = 1; j <= k; j++)
        for (int cj = 0; cj < j; cj++) can[i][ci][j][cj] = 1;
  for (int i = 1; i <= k; i++)
    for (int ci = 0; ci < i; ci++) good[i][ci] = 1;
  for (int v = 0; v < n; v++)
    for (int i = 1; i <= k; i++)
      for (int ci = 0; ci < i; ci++) {
        if (in[v][i][ci] >= 2) good[i][ci] = 0;
        if (in[v][i][ci] > 0)
          for (int j = 1; j < i; j++)
            for (int cj = 0; cj < j; cj++)
              if (in[v][j][cj] > 0) can[i][ci][j][cj] = 0;
      }
  vector<int> c(k + 1);
  int ans = 0;
  function<void(int)> bkt = [&](int v) {
    if (v == k + 1) {
      ans++;
      return;
    }
    c[v] = 0;
    while (c[v] < v) {
      bool ok = good[v][c[v]];
      for (int i = 1; i < v; i++) {
        if (!can[v][c[v]][i][c[i]]) {
          ok = false;
          break;
        }
      }
      if (ok) bkt(v + 1);
      c[v]++;
    }
  };
  bkt(1);
  cout << ans << "\n";
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const int N = 2e5 + 5;
const int K = 10;
bool can[K][K][K][K], good[K][K];
int in[N][K][K];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> from(m), to(m);
  for (int i = 0; i < m; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    x--;
    y--;
    z--;
    from[z] = x;
    to[z] = y;
  }
  vector<vector<int>> out(n);
  for (int i = 0; i < m; i++) out[from[i]].push_back(to[i]);
  for (int i = 0; i < n; i++) {
    int sz = out[i].size();
    for (int j = 0; j < sz; j++) in[out[i][j]][sz][j]++;
  }
  for (int i = 1; i <= k; i++)
    for (int ci = 0; ci < i; ci++)
      for (int j = 1; j <= k; j++)
        for (int cj = 0; cj < j; cj++) can[i][ci][j][cj] = 1;
  for (int i = 1; i <= k; i++)
    for (int ci = 0; ci < i; ci++) good[i][ci] = 1;
  for (int v = 0; v < n; v++)
    for (int i = 1; i <= k; i++)
      for (int ci = 0; ci < i; ci++) {
        if (in[v][i][ci] >= 2) good[i][ci] = 0;
        if (in[v][i][ci] > 0)
          for (int j = 1; j < i; j++)
            for (int cj = 0; cj < j; cj++)
              if (in[v][j][cj] > 0) can[i][ci][j][cj] = 0;
      }
  vector<int> c(k + 1);
  int ans = 0;
  function<void(int)> bkt = [&](int v) {
    if (v == k + 1) {
      ans++;
      return;
    }
    c[v] = 0;
    while (c[v] < v) {
      bool ok = good[v][c[v]];
      for (int i = 1; i < v; i++) {
        if (!can[v][c[v]][i][c[i]]) {
          ok = false;
          break;
        }
      }
      if (ok) bkt(v + 1);
      c[v]++;
    }
  };
  bkt(1);
  cout << ans << "\n";
}

```