#include <bits/stdc++.h>
using namespace std;
const long long logN = 20;
const long long N = 200005;
const long long M = 1000000007;
const long long INF = 1e12;
long long pow1(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % M;
    }
    a = (a * a) % M;
    b >>= 1;
  }
  return res % M;
}
long long n, m, k;
long long ans = 0;
bool bad[10][10];
set<pair<long long, long long>> exe[10][10];
bool mark[10][10];
vector<pair<long long, long long>> tmp[10];
void recur(long long i) {
  if (i > k) {
    ans++;
    return;
  }
  for (long long j = 1; j < i + 1; j++) {
    if (mark[i][j] || bad[i][j]) continue;
    tmp[i].clear();
    for (auto it : exe[i][j]) {
      long long i1 = it.first, c_i1 = it.second;
      if (!mark[i1][c_i1]) {
        mark[i1][c_i1] = 1;
        tmp[i].push_back({i1, c_i1});
      }
    }
    recur(i + 1);
    for (auto it : tmp[i]) mark[it.first][it.second] = 0;
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long T = 1;
  while (T--) {
    cin >> n >> m >> k;
    vector<vector<pair<long long, long long>>> adj(n + 1);
    vector<vector<pair<long long, long long>>> i_ci(n + 1);
    for (long long i = 0; i < m; i++) {
      long long u, v, w;
      cin >> u >> v >> w;
      adj[u].push_back({w, v});
    }
    long long sz = 0;
    for (long long i = 1; i < n + 1; i++) {
      sort(adj[i].begin(), adj[i].end());
    }
    for (long long i = 1; i < n + 1; i++) {
      for (long long j = 0; j < adj[i].size(); j++) {
        i_ci[adj[i][j].second].push_back({adj[i].size(), j + 1});
      }
    }
    set<pair<long long, long long>> st;
    for (long long v = 1; v < n + 1; v++) {
      st.clear();
      auto &tmp = i_ci[v];
      for (auto it : tmp) {
        if (st.count(it)) {
          bad[it.first][it.second] = 1;
          st.erase(it);
        } else {
          if (!bad[it.first][it.second]) st.insert(it);
        }
      }
      for (auto it1 : st) {
        long long i = it1.first, c_i = it1.second;
        for (auto it2 : st) {
          if (i == it2.first && c_i == it2.second) continue;
          long long j = it2.first, c_j = it2.second;
          exe[i][c_i].insert({j, c_j});
        }
      }
    }
    recur(1);
    cout << ans << "\n";
  }
}

### Prompt

Please formulate 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 long long logN = 20;
const long long N = 200005;
const long long M = 1000000007;
const long long INF = 1e12;
long long pow1(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % M;
    }
    a = (a * a) % M;
    b >>= 1;
  }
  return res % M;
}
long long n, m, k;
long long ans = 0;
bool bad[10][10];
set<pair<long long, long long>> exe[10][10];
bool mark[10][10];
vector<pair<long long, long long>> tmp[10];
void recur(long long i) {
  if (i > k) {
    ans++;
    return;
  }
  for (long long j = 1; j < i + 1; j++) {
    if (mark[i][j] || bad[i][j]) continue;
    tmp[i].clear();
    for (auto it : exe[i][j]) {
      long long i1 = it.first, c_i1 = it.second;
      if (!mark[i1][c_i1]) {
        mark[i1][c_i1] = 1;
        tmp[i].push_back({i1, c_i1});
      }
    }
    recur(i + 1);
    for (auto it : tmp[i]) mark[it.first][it.second] = 0;
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long T = 1;
  while (T--) {
    cin >> n >> m >> k;
    vector<vector<pair<long long, long long>>> adj(n + 1);
    vector<vector<pair<long long, long long>>> i_ci(n + 1);
    for (long long i = 0; i < m; i++) {
      long long u, v, w;
      cin >> u >> v >> w;
      adj[u].push_back({w, v});
    }
    long long sz = 0;
    for (long long i = 1; i < n + 1; i++) {
      sort(adj[i].begin(), adj[i].end());
    }
    for (long long i = 1; i < n + 1; i++) {
      for (long long j = 0; j < adj[i].size(); j++) {
        i_ci[adj[i][j].second].push_back({adj[i].size(), j + 1});
      }
    }
    set<pair<long long, long long>> st;
    for (long long v = 1; v < n + 1; v++) {
      st.clear();
      auto &tmp = i_ci[v];
      for (auto it : tmp) {
        if (st.count(it)) {
          bad[it.first][it.second] = 1;
          st.erase(it);
        } else {
          if (!bad[it.first][it.second]) st.insert(it);
        }
      }
      for (auto it1 : st) {
        long long i = it1.first, c_i = it1.second;
        for (auto it2 : st) {
          if (i == it2.first && c_i == it2.second) continue;
          long long j = it2.first, c_j = it2.second;
          exe[i][c_i].insert({j, c_j});
        }
      }
    }
    recur(1);
    cout << ans << "\n";
  }
}

```