#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
vector<vector<pair<int, int>>> adj(222222);
vector<long long int> val(222222);
long long int MOD1 = 1e9 + 7;
long long int MOD2 = 998244353;
struct Hash {
  long long int h[2];
  void make(vector<int> a) {
    h[0] = h[1] = 1;
    for (long long int x : a) {
      h[0] = h[0] * val[x] % MOD1;
      h[1] = h[1] * val[x] % MOD2;
    }
  }
  Hash() {}
  bool operator==(const Hash &h1) { return h1.h[0] == h[0] && h1.h[1] == h[1]; }
  Hash operator*(const Hash &h1) {
    Hash res;
    res.h[0] = h[0];
    res.h[1] = h[1];
    res.h[0] = res.h[0] * h1.h[0] % MOD1;
    res.h[1] = res.h[1] * h1.h[1] % MOD2;
    return res;
  }
} tar;
int n, m, k;
long long int ans = 0;
vector<vector<Hash>> h(11, vector<Hash>(11));
void dfs(int i, Hash cur) {
  if (i == k + 1) {
    if (cur == tar) ans++;
    return;
  }
  for (int j = 1; j <= i; j++) {
    Hash tmp;
    tmp.h[0] = cur.h[0];
    tmp.h[1] = cur.h[1];
    tmp = tmp * h[i][j];
    dfs(i + 1, tmp);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++)
    val[i] = uniform_int_distribution<long long int>(1e7, 9e8 + 6)(rng);
  for (int i = 0; i < m; i++) {
    int v, u, w;
    cin >> v >> u >> w;
    adj[v].push_back({w, u});
  }
  for (int i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end());
  vector<vector<vector<int>>> s(k + 1, vector<vector<int>>(k + 1));
  for (int i = 1; i <= n; i++) {
    int d = (int)(adj[i].size());
    for (int j = 0; j < (int)(adj[i].size()); j++) {
      s[d][j + 1].push_back(adj[i][j].second);
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      h[i][j].make(s[i][j]);
    }
  }
  vector<int> a;
  for (int i = 1; i <= n; i++) a.push_back(i);
  tar.make(a);
  Hash b;
  a.clear();
  b.make(a);
  dfs(1, b);
  cout << ans;
}

### Prompt

Please create a solution in Cpp 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("Ofast")
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
vector<vector<pair<int, int>>> adj(222222);
vector<long long int> val(222222);
long long int MOD1 = 1e9 + 7;
long long int MOD2 = 998244353;
struct Hash {
  long long int h[2];
  void make(vector<int> a) {
    h[0] = h[1] = 1;
    for (long long int x : a) {
      h[0] = h[0] * val[x] % MOD1;
      h[1] = h[1] * val[x] % MOD2;
    }
  }
  Hash() {}
  bool operator==(const Hash &h1) { return h1.h[0] == h[0] && h1.h[1] == h[1]; }
  Hash operator*(const Hash &h1) {
    Hash res;
    res.h[0] = h[0];
    res.h[1] = h[1];
    res.h[0] = res.h[0] * h1.h[0] % MOD1;
    res.h[1] = res.h[1] * h1.h[1] % MOD2;
    return res;
  }
} tar;
int n, m, k;
long long int ans = 0;
vector<vector<Hash>> h(11, vector<Hash>(11));
void dfs(int i, Hash cur) {
  if (i == k + 1) {
    if (cur == tar) ans++;
    return;
  }
  for (int j = 1; j <= i; j++) {
    Hash tmp;
    tmp.h[0] = cur.h[0];
    tmp.h[1] = cur.h[1];
    tmp = tmp * h[i][j];
    dfs(i + 1, tmp);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++)
    val[i] = uniform_int_distribution<long long int>(1e7, 9e8 + 6)(rng);
  for (int i = 0; i < m; i++) {
    int v, u, w;
    cin >> v >> u >> w;
    adj[v].push_back({w, u});
  }
  for (int i = 1; i <= n; i++) sort(adj[i].begin(), adj[i].end());
  vector<vector<vector<int>>> s(k + 1, vector<vector<int>>(k + 1));
  for (int i = 1; i <= n; i++) {
    int d = (int)(adj[i].size());
    for (int j = 0; j < (int)(adj[i].size()); j++) {
      s[d][j + 1].push_back(adj[i][j].second);
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      h[i][j].make(s[i][j]);
    }
  }
  vector<int> a;
  for (int i = 1; i <= n; i++) a.push_back(i);
  tar.make(a);
  Hash b;
  a.clear();
  b.make(a);
  dfs(1, b);
  cout << ans;
}

```