#include <bits/stdc++.h>
using namespace std;
const int N = 200006;
vector<pair<int, int>> E[N];
int n, m, k, ans = 0;
bitset<N> b[10][10];
vector<vector<bool>> good(10, vector<bool>(10, true));
bool can[10][10][10][10];
void go(vector<int> now) {
  if (now.size() == k) {
    ans += 1;
    return;
  }
  int cur = now.size() + 1;
  for (int i = 0; i < cur; i++) {
    if (good[cur][i]) {
      bool B = true;
      for (int j = 1; j < cur; j++) {
        B &= (can[cur][i][j][now[j - 1]]);
      }
      if (B) {
        now.push_back(i);
        go(now);
        now.pop_back();
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int x, y, cost;
    cin >> x >> y >> cost;
    E[x].push_back({cost, y});
  }
  for (int deg = 1; deg <= k; deg++) {
    for (int i = 1; i <= n; i++) {
      if (E[i].size() == deg) {
        sort(E[i].begin(), E[i].end());
        for (int j = 0; j < E[i].size(); j++) {
          int neig = E[i][j].second;
          if (b[deg][j][neig]) {
            good[deg][j] = false;
          }
          b[deg][j][neig] = 1;
        }
      }
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int x = 0; x < i; x++) {
      for (int j = 1; j <= k; j++) {
        for (int y = 0; y < j; y++) {
          if (good[i][x] && good[j][y] && (b[i][x] & b[j][y]).count() == 0) {
            can[i][x][j][y] = can[j][y][i][x] = true;
          }
        }
      }
    }
  }
  go({});
  cout << ans << '\n';
  return 0;
}

### Prompt

Generate 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>
using namespace std;
const int N = 200006;
vector<pair<int, int>> E[N];
int n, m, k, ans = 0;
bitset<N> b[10][10];
vector<vector<bool>> good(10, vector<bool>(10, true));
bool can[10][10][10][10];
void go(vector<int> now) {
  if (now.size() == k) {
    ans += 1;
    return;
  }
  int cur = now.size() + 1;
  for (int i = 0; i < cur; i++) {
    if (good[cur][i]) {
      bool B = true;
      for (int j = 1; j < cur; j++) {
        B &= (can[cur][i][j][now[j - 1]]);
      }
      if (B) {
        now.push_back(i);
        go(now);
        now.pop_back();
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int x, y, cost;
    cin >> x >> y >> cost;
    E[x].push_back({cost, y});
  }
  for (int deg = 1; deg <= k; deg++) {
    for (int i = 1; i <= n; i++) {
      if (E[i].size() == deg) {
        sort(E[i].begin(), E[i].end());
        for (int j = 0; j < E[i].size(); j++) {
          int neig = E[i][j].second;
          if (b[deg][j][neig]) {
            good[deg][j] = false;
          }
          b[deg][j][neig] = 1;
        }
      }
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int x = 0; x < i; x++) {
      for (int j = 1; j <= k; j++) {
        for (int y = 0; y < j; y++) {
          if (good[i][x] && good[j][y] && (b[i][x] & b[j][y]).count() == 0) {
            can[i][x][j][y] = can[j][y][i][x] = true;
          }
        }
      }
    }
  }
  go({});
  cout << ans << '\n';
  return 0;
}

```