#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void chmin(T &a, const T &val) {
  if (a > val) a = val;
}
template <class T>
inline void chmax(T &a, const T &val) {
  if (a < val) a = val;
}
template <class T>
long long sqr(T a) {
  return 1ll * a * a;
}
template <class T>
void compress(T &a) {
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
}
const double pi = acos(-1.00);
const double eps = 1e-6;
const int INF = 1e9;
const int MOD = 1e9 + 7;
const int dx[4] = {0, 0, -1, 1};
const int dy[4] = {-1, 1, 0, 0};
const int N = 2e5 + 1;
int n, m, k, ans, deg[N];
long long mask[10][10], P[N], base;
vector<pair<int, int> > a[N];
void Try(int i, long long pre = 0) {
  for (int j = 1; j <= i; j++) {
    long long cur = (pre + mask[i][j]) % MOD;
    if (i < k)
      Try(i + 1, cur);
    else {
      ans += (cur == base);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  P[0] = 1;
  for (int i = 1; i <= n; i++) P[i] = P[i - 1] * 234567 % MOD;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    a[u].push_back({v, w});
    deg[u]++;
  }
  for (int u = 1; u <= n; u++) {
    sort(a[u].begin(), a[u].end(), [&](pair<int, int> x, pair<int, int> y) {
      return x.second < y.second;
    });
    for (int i = 0; i < a[u].size(); i++) {
      int v = a[u][i].first;
      (mask[deg[u]][i + 1] += 1ll * v * P[v - 1]) %= MOD;
    }
  }
  for (int i = 1; i <= n; i++) base = (base + 1ll * i * P[i - 1]) % MOD;
  Try(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 namespace std;
template <class T>
inline void chmin(T &a, const T &val) {
  if (a > val) a = val;
}
template <class T>
inline void chmax(T &a, const T &val) {
  if (a < val) a = val;
}
template <class T>
long long sqr(T a) {
  return 1ll * a * a;
}
template <class T>
void compress(T &a) {
  sort(a.begin(), a.end());
  a.resize(unique(a.begin(), a.end()) - a.begin());
}
const double pi = acos(-1.00);
const double eps = 1e-6;
const int INF = 1e9;
const int MOD = 1e9 + 7;
const int dx[4] = {0, 0, -1, 1};
const int dy[4] = {-1, 1, 0, 0};
const int N = 2e5 + 1;
int n, m, k, ans, deg[N];
long long mask[10][10], P[N], base;
vector<pair<int, int> > a[N];
void Try(int i, long long pre = 0) {
  for (int j = 1; j <= i; j++) {
    long long cur = (pre + mask[i][j]) % MOD;
    if (i < k)
      Try(i + 1, cur);
    else {
      ans += (cur == base);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  P[0] = 1;
  for (int i = 1; i <= n; i++) P[i] = P[i - 1] * 234567 % MOD;
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    a[u].push_back({v, w});
    deg[u]++;
  }
  for (int u = 1; u <= n; u++) {
    sort(a[u].begin(), a[u].end(), [&](pair<int, int> x, pair<int, int> y) {
      return x.second < y.second;
    });
    for (int i = 0; i < a[u].size(); i++) {
      int v = a[u][i].first;
      (mask[deg[u]][i + 1] += 1ll * v * P[v - 1]) %= MOD;
    }
  }
  for (int i = 1; i <= n; i++) base = (base + 1ll * i * P[i - 1]) % MOD;
  Try(1);
  cout << ans;
  return 0;
}

```