#include <bits/stdc++.h>
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using namespace std;
const int N = 2e5 + 2, mod = 1e9 + 7;
int n, m, k, pw[N], sum[10][10], arr[10], tot, ans, hs;
vector<pair<int, int> > g[N];
vector<int> versz[10];
void permute(int i) {
  if (i == k + 1) {
    ans += (hs == tot);
    return;
  }
  for (int j = (1); j <= (i); ++j) {
    arr[i] = j;
    hs = (hs + sum[i][j]) % mod;
    permute(i + 1);
    hs = (hs - sum[i][j] + mod) % mod;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m >> k;
  for (int i = (1); i <= (m); ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    g[u].emplace_back(v, w);
  }
  for (int i = (1); i <= (n); ++i) {
    sort((g[i]).begin(), (g[i]).end(),
         [&](const pair<int, int>& A, const pair<int, int>& B) {
           return A.second < B.second;
         });
    versz[((int)(g[i]).size())].emplace_back(i);
  }
  tot = 0;
  pw[0] = 1;
  for (int i = (1); i <= (n); ++i)
    pw[i] = pw[i - 1] * 1LL * 31 % mod, tot = (tot + pw[i]) % mod;
  for (int i = (1); i <= (k); ++i) {
    for (int j = (1); j <= (i); ++j)
      for (int v : versz[i])
        sum[i][j] = (sum[i][j] + pw[g[v][j - 1].first]) % mod;
  }
  permute(1);
  cout << ans;
  return 0;
}

### Prompt

Generate 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 ll = long long;
using ld = long double;
using ull = unsigned long long;
using namespace std;
const int N = 2e5 + 2, mod = 1e9 + 7;
int n, m, k, pw[N], sum[10][10], arr[10], tot, ans, hs;
vector<pair<int, int> > g[N];
vector<int> versz[10];
void permute(int i) {
  if (i == k + 1) {
    ans += (hs == tot);
    return;
  }
  for (int j = (1); j <= (i); ++j) {
    arr[i] = j;
    hs = (hs + sum[i][j]) % mod;
    permute(i + 1);
    hs = (hs - sum[i][j] + mod) % mod;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m >> k;
  for (int i = (1); i <= (m); ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    g[u].emplace_back(v, w);
  }
  for (int i = (1); i <= (n); ++i) {
    sort((g[i]).begin(), (g[i]).end(),
         [&](const pair<int, int>& A, const pair<int, int>& B) {
           return A.second < B.second;
         });
    versz[((int)(g[i]).size())].emplace_back(i);
  }
  tot = 0;
  pw[0] = 1;
  for (int i = (1); i <= (n); ++i)
    pw[i] = pw[i - 1] * 1LL * 31 % mod, tot = (tot + pw[i]) % mod;
  for (int i = (1); i <= (k); ++i) {
    for (int j = (1); j <= (i); ++j)
      for (int v : versz[i])
        sum[i][j] = (sum[i][j] + pw[g[v][j - 1].first]) % mod;
  }
  permute(1);
  cout << ans;
  return 0;
}

```