#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 5;
const int P = 998244353;
const int inf = 0x3f3f3f3f;
bitset<maxn> is[10][10];
int n, m, k, c[10], ans, vis[maxn], s[maxn], cnt[10][10], jd[10][10][10][10];
vector<pair<int, int>> G[maxn];
inline int read() {
  char ch = getchar();
  int u = 0, f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    u = u * 10 + ch - 48;
    ch = getchar();
  }
  return u * f;
}
inline bool ok(int a, int b) {
  for (int i = 1; i <= a - 1; ++i)
    if (jd[a][b][i][c[i]]) return false;
  return true;
}
inline int dfs(int x, int num, int rnt = 0) {
  if (x == k + 1) return num == n;
  for (int i = 1; i <= x; ++i) {
    if (ok(x, i)) {
      c[x] = i;
      rnt += dfs(x + 1, num + cnt[x][i]);
    }
  }
  return rnt;
}
int main() {
  n = read();
  m = read();
  k = read();
  for (int i = 1; i <= m; ++i) {
    int x = read(), y = read(), w = read();
    G[x].push_back(make_pair(w, y));
  }
  for (int i = 1; i <= n; ++i) {
    sort(G[i].begin(), G[i].end());
    s[i] = ((int)G[i].size());
    for (int j = 1; j <= s[i]; ++j) is[s[i]][j].set(G[i][j - 1].second);
  }
  for (int i = 1; i <= k; ++i)
    for (int j = 1; j <= i; ++j) cnt[i][j] = is[i][j].count();
  for (int a = 1; a <= k; ++a)
    for (int b = 1; b <= a; ++b)
      for (int c = 1; c <= k; ++c)
        for (int d = 1; d <= c; ++d)
          if ((is[a][b] & is[c][d]).count()) jd[a][b][c][d] = 1;
  cout << dfs(1, 0) << endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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 maxn = 2e5 + 5;
const int P = 998244353;
const int inf = 0x3f3f3f3f;
bitset<maxn> is[10][10];
int n, m, k, c[10], ans, vis[maxn], s[maxn], cnt[10][10], jd[10][10][10][10];
vector<pair<int, int>> G[maxn];
inline int read() {
  char ch = getchar();
  int u = 0, f = 1;
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    u = u * 10 + ch - 48;
    ch = getchar();
  }
  return u * f;
}
inline bool ok(int a, int b) {
  for (int i = 1; i <= a - 1; ++i)
    if (jd[a][b][i][c[i]]) return false;
  return true;
}
inline int dfs(int x, int num, int rnt = 0) {
  if (x == k + 1) return num == n;
  for (int i = 1; i <= x; ++i) {
    if (ok(x, i)) {
      c[x] = i;
      rnt += dfs(x + 1, num + cnt[x][i]);
    }
  }
  return rnt;
}
int main() {
  n = read();
  m = read();
  k = read();
  for (int i = 1; i <= m; ++i) {
    int x = read(), y = read(), w = read();
    G[x].push_back(make_pair(w, y));
  }
  for (int i = 1; i <= n; ++i) {
    sort(G[i].begin(), G[i].end());
    s[i] = ((int)G[i].size());
    for (int j = 1; j <= s[i]; ++j) is[s[i]][j].set(G[i][j - 1].second);
  }
  for (int i = 1; i <= k; ++i)
    for (int j = 1; j <= i; ++j) cnt[i][j] = is[i][j].count();
  for (int a = 1; a <= k; ++a)
    for (int b = 1; b <= a; ++b)
      for (int c = 1; c <= k; ++c)
        for (int d = 1; d <= c; ++d)
          if ((is[a][b] & is[c][d]).count()) jd[a][b][c][d] = 1;
  cout << dfs(1, 0) << endl;
  return 0;
}

```