#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int gcd(int x, int y) {
  if (!y)
    return x;
  else
    return gcd(y, x % y);
}
struct Edge {
  int v, w;
  bool operator<(const Edge &x) const { return w < x.w; }
};
vector<Edge> e[N];
int ans;
int n;
int a[15];
int b[N][15][15];
int c[15][15];
int d[15][15][15][15];
void dfs(int u, int k) {
  if (u == k + 1) {
    for (int i = 1; i <= k; i++)
      for (int j = i + 1; j <= k; j++) {
        if (d[i][a[i]][j][a[j]]) return;
      }
    ans++;
    return;
  }
  for (int i = 0; i < u; i++) {
    if (c[u][i]) continue;
    a[u] = i, dfs(u + 1, k);
  }
}
int main() {
  int m, u, v, w, k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &u, &v, &w);
    e[u].push_back({v, w});
  }
  for (int i = 1; i <= n; i++) {
    sort(e[i].begin(), e[i].end());
    for (int j = 0; j < e[i].size(); j++) b[e[i][j].v][e[i].size()][j]++;
  }
  for (int u = 1; u <= n; u++) {
    vector<pair<int, int> > vec;
    for (int i = 1; i <= k; i++)
      for (int j = 0; j <= i - 1; j++) {
        if (b[u][i][j] > 1)
          c[i][j] = 1;
        else if (b[u][i][j] == 1)
          vec.push_back({i, j});
      }
    for (int i = 0; i < vec.size(); i++)
      for (int j = 1; j < vec.size(); j++)
        d[vec[i].first][vec[i].second][vec[j].first][vec[j].second] = 1;
  }
  dfs(1, k);
  printf("%d\n", ans);
  return 0;
}

### 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 int N = 2e5 + 5;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int gcd(int x, int y) {
  if (!y)
    return x;
  else
    return gcd(y, x % y);
}
struct Edge {
  int v, w;
  bool operator<(const Edge &x) const { return w < x.w; }
};
vector<Edge> e[N];
int ans;
int n;
int a[15];
int b[N][15][15];
int c[15][15];
int d[15][15][15][15];
void dfs(int u, int k) {
  if (u == k + 1) {
    for (int i = 1; i <= k; i++)
      for (int j = i + 1; j <= k; j++) {
        if (d[i][a[i]][j][a[j]]) return;
      }
    ans++;
    return;
  }
  for (int i = 0; i < u; i++) {
    if (c[u][i]) continue;
    a[u] = i, dfs(u + 1, k);
  }
}
int main() {
  int m, u, v, w, k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= m; i++) {
    scanf("%d%d%d", &u, &v, &w);
    e[u].push_back({v, w});
  }
  for (int i = 1; i <= n; i++) {
    sort(e[i].begin(), e[i].end());
    for (int j = 0; j < e[i].size(); j++) b[e[i][j].v][e[i].size()][j]++;
  }
  for (int u = 1; u <= n; u++) {
    vector<pair<int, int> > vec;
    for (int i = 1; i <= k; i++)
      for (int j = 0; j <= i - 1; j++) {
        if (b[u][i][j] > 1)
          c[i][j] = 1;
        else if (b[u][i][j] == 1)
          vec.push_back({i, j});
      }
    for (int i = 0; i < vec.size(); i++)
      for (int j = 1; j < vec.size(); j++)
        d[vec[i].first][vec[i].second][vec[j].first][vec[j].second] = 1;
  }
  dfs(1, k);
  printf("%d\n", ans);
  return 0;
}

```