#include <bits/stdc++.h>
#pragma GCC optimize 2
using namespace std;
vector<pair<int, int> > e[200005];
bitset<200005> s[10][10];
int D[200005];
int pd[10][10][10][10];
int ans = 0, p[10], k;
inline void dfs(int x) {
  if (x == k + 1) {
    ++ans;
    return;
  }
  for (int i = 1; i <= x; i++) {
    int flag = 1;
    if (s[x][i][200001]) continue;
    for (int j = 1; j < x; j++) {
      if (!pd[x][i][j][p[j]]) {
        flag = 0;
        break;
      }
    }
    if (flag) p[x] = i, dfs(x + 1);
  }
}
int main(int argc, char** argv) {
  int n, m;
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    e[u].push_back({w, v});
    ++D[u];
  }
  for (int i = 1; i <= n; i++) {
    sort(e[i].begin(), e[i].end());
    for (int j = 0; j < e[i].size(); j++) {
      if (s[D[i]][j + 1][e[i][j].second] && !s[D[i]][j + 1][200001])
        s[D[i]][j + 1][200001] = 1;
      s[D[i]][j + 1][e[i][j].second] = 1;
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      for (int x = 1; x < i; x++) {
        for (int y = 1; y <= x; y++) {
          if ((s[i][j] & s[x][y]).none()) {
            pd[i][j][x][y] = 1;
          }
        }
      }
    }
  }
  dfs(1);
  cout << ans;
  return 0;
}

### Prompt

Your task is to create a cpp solution 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>
#pragma GCC optimize 2
using namespace std;
vector<pair<int, int> > e[200005];
bitset<200005> s[10][10];
int D[200005];
int pd[10][10][10][10];
int ans = 0, p[10], k;
inline void dfs(int x) {
  if (x == k + 1) {
    ++ans;
    return;
  }
  for (int i = 1; i <= x; i++) {
    int flag = 1;
    if (s[x][i][200001]) continue;
    for (int j = 1; j < x; j++) {
      if (!pd[x][i][j][p[j]]) {
        flag = 0;
        break;
      }
    }
    if (flag) p[x] = i, dfs(x + 1);
  }
}
int main(int argc, char** argv) {
  int n, m;
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    e[u].push_back({w, v});
    ++D[u];
  }
  for (int i = 1; i <= n; i++) {
    sort(e[i].begin(), e[i].end());
    for (int j = 0; j < e[i].size(); j++) {
      if (s[D[i]][j + 1][e[i][j].second] && !s[D[i]][j + 1][200001])
        s[D[i]][j + 1][200001] = 1;
      s[D[i]][j + 1][e[i][j].second] = 1;
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      for (int x = 1; x < i; x++) {
        for (int y = 1; y <= x; y++) {
          if ((s[i][j] & s[x][y]).none()) {
            pd[i][j][x][y] = 1;
          }
        }
      }
    }
  }
  dfs(1);
  cout << ans;
  return 0;
}

```