#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int NAX = 2e5 + 5, MOD = 1000000007;
const int HS = 3, K = 10;
int mod[HS];
struct Hash {
  ll a[HS];
  Hash() {
    fill(a, a + HS, 0);
    a[1] = 1;
  }
  Hash(ll x) {
    for (size_t i = 0; i < HS; i++) a[i] = (x % mod[i] + mod[i]) % mod[i];
  }
  Hash operator+(Hash rhs) {
    Hash res;
    res.a[0] = (a[0] + rhs.a[0]) % mod[0];
    res.a[1] = (a[1] * rhs.a[1]) % mod[1];
    res.a[2] = (a[2] + rhs.a[2]) % mod[2];
    return res;
  }
  bool operator==(const Hash &x) const {
    for (size_t i = 0; i < HS; i++)
      if (a[i] != x.a[i]) return false;
    return true;
  }
} val[NAX], c[K][K], second;
ostream &operator<<(ostream &ou, const Hash &hs) {
  ou << hs.a[0] << ' ' << hs.a[1] << ' ' << hs.a[2] << '\n';
  return ou;
};
int n, m, k;
int ans = 0;
vector<pair<int, int>> adj[NAX];
void dfs(int x, Hash hsh) {
  if (x == k) {
    ;
    if (hsh == second) ans++;
    return;
  }
  for (size_t i = 1; i <= (x + 1); i++) dfs(x + 1, hsh + c[x + 1][i]);
}
struct Solution {
  Solution() {}
  void solveCase() {
    srand(time(NULL));
    mod[0] = MOD;
    mod[1] = 1e9 + 9;
    mod[2] = 1e8 + rand() % int(1e8) + 1;
    ;
    cin >> n >> m >> k;
    for (size_t i = 0; i < m; i++) {
      int u, v, w;
      cin >> u >> v >> w;
      adj[u].push_back({w, v});
    }
    for (size_t i = 1; i <= n; i++) {
      val[i] = Hash(rand() % MOD + 1);
      second = second + val[i];
    }
    for (size_t i = 1; i <= n; i++) {
      int d = adj[i].size();
      sort(adj[i].begin(), adj[i].end());
      for (int j = 1; j <= d; j++) {
        int v = adj[i][j - 1].second;
        c[d][j] = c[d][j] + val[v];
      }
    }
    dfs(0, Hash());
    cout << ans << '\n';
  }
};
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  Solution mySolver;
  for (int i = 1; i <= t; ++i) {
    mySolver.solveCase();
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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;
using ll = long long;
const int NAX = 2e5 + 5, MOD = 1000000007;
const int HS = 3, K = 10;
int mod[HS];
struct Hash {
  ll a[HS];
  Hash() {
    fill(a, a + HS, 0);
    a[1] = 1;
  }
  Hash(ll x) {
    for (size_t i = 0; i < HS; i++) a[i] = (x % mod[i] + mod[i]) % mod[i];
  }
  Hash operator+(Hash rhs) {
    Hash res;
    res.a[0] = (a[0] + rhs.a[0]) % mod[0];
    res.a[1] = (a[1] * rhs.a[1]) % mod[1];
    res.a[2] = (a[2] + rhs.a[2]) % mod[2];
    return res;
  }
  bool operator==(const Hash &x) const {
    for (size_t i = 0; i < HS; i++)
      if (a[i] != x.a[i]) return false;
    return true;
  }
} val[NAX], c[K][K], second;
ostream &operator<<(ostream &ou, const Hash &hs) {
  ou << hs.a[0] << ' ' << hs.a[1] << ' ' << hs.a[2] << '\n';
  return ou;
};
int n, m, k;
int ans = 0;
vector<pair<int, int>> adj[NAX];
void dfs(int x, Hash hsh) {
  if (x == k) {
    ;
    if (hsh == second) ans++;
    return;
  }
  for (size_t i = 1; i <= (x + 1); i++) dfs(x + 1, hsh + c[x + 1][i]);
}
struct Solution {
  Solution() {}
  void solveCase() {
    srand(time(NULL));
    mod[0] = MOD;
    mod[1] = 1e9 + 9;
    mod[2] = 1e8 + rand() % int(1e8) + 1;
    ;
    cin >> n >> m >> k;
    for (size_t i = 0; i < m; i++) {
      int u, v, w;
      cin >> u >> v >> w;
      adj[u].push_back({w, v});
    }
    for (size_t i = 1; i <= n; i++) {
      val[i] = Hash(rand() % MOD + 1);
      second = second + val[i];
    }
    for (size_t i = 1; i <= n; i++) {
      int d = adj[i].size();
      sort(adj[i].begin(), adj[i].end());
      for (int j = 1; j <= d; j++) {
        int v = adj[i][j - 1].second;
        c[d][j] = c[d][j] + val[v];
      }
    }
    dfs(0, Hash());
    cout << ans << '\n';
  }
};
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  Solution mySolver;
  for (int i = 1; i <= t; ++i) {
    mySolver.solveCase();
  }
  return 0;
}

```