#include <bits/stdc++.h>
using namespace std;
int n, m, k;
int const K = 9;
int const N = 2e5;
int const num_hash = 3;
mt19937 mt_rand(
    chrono::high_resolution_clock::now().time_since_epoch().count());
long long mod[num_hash];
struct hash_number {
  long long a[num_hash];
  hash_number() { fill(a, a + num_hash, 0); }
  hash_number(long long x) {
    for (int i = (0); i < (num_hash); ++i) a[i] = x % mod[i];
  }
  hash_number operator+(hash_number x) {
    hash_number res;
    res.a[0] = (a[0] + x.a[0]) % mod[0];
    res.a[1] = (a[1] * (long long)1ll * x.a[1]) % mod[1];
    res.a[2] = (a[2] + x.a[2]) % mod[2];
    return res;
  }
  bool operator==(const hash_number& x) const {
    for (int i = (0); i < (num_hash); ++i)
      if (a[i] != x.a[i]) return 0;
    return 1;
  }
} val[N], hashes[K + 1][K], ans;
int recur(int now, int cho, hash_number h) {
  h = h + hashes[now][cho];
  if (now == k) {
    if (h == ans) {
      return 1;
    } else
      return 0;
  }
  int cnt = 0;
  for (int i = (0); i < (now + 1); ++i) {
    cnt += recur(now + 1, i, h);
  }
  return cnt;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  int u, v, w;
  int out[n];
  fill(out, out + n, 0);
  vector<pair<int, int>> adj[n];
  for (int i = (0); i < (m); ++i) {
    cin >> u >> v >> w;
    --u;
    --v;
    adj[u].push_back({w, v});
    ++out[u];
  }
  uniform_int_distribution<long long> rg(1, 1e18);
  mod[0] = 998244353;
  mod[1] = 1e9 + 7;
  mod[2] = uniform_int_distribution<int>(1e8, 1e9)(mt_rand);
  for (int i = (0); i < (n + 1); ++i) val[i] = hash_number(rg(mt_rand));
  for (int i = (0); i < (n); ++i) {
    sort(begin(adj[i]), end(adj[i]));
    for (int j = (0); j < (out[i]); ++j) {
      hashes[out[i]][j] = hashes[out[i]][j] + val[adj[i][j].second];
    }
  }
  for (int i = (0); i < (n); ++i) {
    ans = ans + val[i];
  }
  cout << recur(1, 0, hash_number()) << '\n';
  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;
int n, m, k;
int const K = 9;
int const N = 2e5;
int const num_hash = 3;
mt19937 mt_rand(
    chrono::high_resolution_clock::now().time_since_epoch().count());
long long mod[num_hash];
struct hash_number {
  long long a[num_hash];
  hash_number() { fill(a, a + num_hash, 0); }
  hash_number(long long x) {
    for (int i = (0); i < (num_hash); ++i) a[i] = x % mod[i];
  }
  hash_number operator+(hash_number x) {
    hash_number res;
    res.a[0] = (a[0] + x.a[0]) % mod[0];
    res.a[1] = (a[1] * (long long)1ll * x.a[1]) % mod[1];
    res.a[2] = (a[2] + x.a[2]) % mod[2];
    return res;
  }
  bool operator==(const hash_number& x) const {
    for (int i = (0); i < (num_hash); ++i)
      if (a[i] != x.a[i]) return 0;
    return 1;
  }
} val[N], hashes[K + 1][K], ans;
int recur(int now, int cho, hash_number h) {
  h = h + hashes[now][cho];
  if (now == k) {
    if (h == ans) {
      return 1;
    } else
      return 0;
  }
  int cnt = 0;
  for (int i = (0); i < (now + 1); ++i) {
    cnt += recur(now + 1, i, h);
  }
  return cnt;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  int u, v, w;
  int out[n];
  fill(out, out + n, 0);
  vector<pair<int, int>> adj[n];
  for (int i = (0); i < (m); ++i) {
    cin >> u >> v >> w;
    --u;
    --v;
    adj[u].push_back({w, v});
    ++out[u];
  }
  uniform_int_distribution<long long> rg(1, 1e18);
  mod[0] = 998244353;
  mod[1] = 1e9 + 7;
  mod[2] = uniform_int_distribution<int>(1e8, 1e9)(mt_rand);
  for (int i = (0); i < (n + 1); ++i) val[i] = hash_number(rg(mt_rand));
  for (int i = (0); i < (n); ++i) {
    sort(begin(adj[i]), end(adj[i]));
    for (int j = (0); j < (out[i]); ++j) {
      hashes[out[i]][j] = hashes[out[i]][j] + val[adj[i][j].second];
    }
  }
  for (int i = (0); i < (n); ++i) {
    ans = ans + val[i];
  }
  cout << recur(1, 0, hash_number()) << '\n';
  return 0;
}

```