#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T& x) {
  x = 0;
  bool f = 0;
  char ch = (char)getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 1;
    ch = (char)getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch & 15);
    ch = (char)getchar();
  }
  if (f) x = -x;
}
template <typename T>
inline void write(T x) {
  static int st[35];
  int top = 0;
  do {
    st[top++] = x % 10;
    x /= 10;
  } while (x);
  while (top) putchar(st[--top] + '0');
}
const double pi = acos(-1.0);
const double eps = (double)1e-4;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 5;
const int maxm = 2e5 + 10;
const long long p = 31;
int n, m, k;
vector<pair<int, int> > adj[maxn];
int out[maxn];
vector<int> deg[10];
long long pow31[maxn];
long long ha[10][10];
long long tar = 0;
int ans = 0;
void debug() {
  cout << "tar = " << tar << endl;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      printf("ha[%d][%d] = %lld\n", i, j, ha[i][j]);
    }
  }
}
void init() {
  pow31[0] = 1;
  for (long long i = 1; i <= n; i++) pow31[i] = (pow31[i - 1] * p) % mod;
  for (long long i = 1; i <= n; i++) tar += pow31[i];
  tar %= mod;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      for (int a : deg[i]) {
        int to = adj[a][j - 1].second;
        (ha[i][j] += pow31[to]) %= mod;
      }
    }
  }
}
void dfs(int dep, long long cur) {
  if (dep == k + 1) {
    if (tar == cur) ans++;
    return;
  }
  for (int j = 1; j <= dep; j++) {
    dfs(dep + 1, (cur + ha[dep][j]) % mod);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    adj[u].push_back({w, v});
    out[u]++;
  }
  for (int i = 1; i <= n; i++) {
    sort(adj[i].begin(), adj[i].end());
    deg[out[i]].push_back(i);
  }
  init();
  dfs(1, 0);
  cout << ans << endl;
}

### 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;
template <typename T>
inline void read(T& x) {
  x = 0;
  bool f = 0;
  char ch = (char)getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 1;
    ch = (char)getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch & 15);
    ch = (char)getchar();
  }
  if (f) x = -x;
}
template <typename T>
inline void write(T x) {
  static int st[35];
  int top = 0;
  do {
    st[top++] = x % 10;
    x /= 10;
  } while (x);
  while (top) putchar(st[--top] + '0');
}
const double pi = acos(-1.0);
const double eps = (double)1e-4;
const int mod = 1e9 + 7;
const int maxn = 2e5 + 5;
const int maxm = 2e5 + 10;
const long long p = 31;
int n, m, k;
vector<pair<int, int> > adj[maxn];
int out[maxn];
vector<int> deg[10];
long long pow31[maxn];
long long ha[10][10];
long long tar = 0;
int ans = 0;
void debug() {
  cout << "tar = " << tar << endl;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      printf("ha[%d][%d] = %lld\n", i, j, ha[i][j]);
    }
  }
}
void init() {
  pow31[0] = 1;
  for (long long i = 1; i <= n; i++) pow31[i] = (pow31[i - 1] * p) % mod;
  for (long long i = 1; i <= n; i++) tar += pow31[i];
  tar %= mod;
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      for (int a : deg[i]) {
        int to = adj[a][j - 1].second;
        (ha[i][j] += pow31[to]) %= mod;
      }
    }
  }
}
void dfs(int dep, long long cur) {
  if (dep == k + 1) {
    if (tar == cur) ans++;
    return;
  }
  for (int j = 1; j <= dep; j++) {
    dfs(dep + 1, (cur + ha[dep][j]) % mod);
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    adj[u].push_back({w, v});
    out[u]++;
  }
  for (int i = 1; i <= n; i++) {
    sort(adj[i].begin(), adj[i].end());
    deg[out[i]].push_back(i);
  }
  init();
  dfs(1, 0);
  cout << ans << endl;
}

```