#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
template <typename T>
using pxx = pair<T, T>;
ll power(ll x, ll y);
const ll sp1 = 1e12 + 39;
const ll sp2 = 1e15 + 1e5 + 11;
const ll INF = 1e18 + 18;
const ll MOD = 1e9 + 7;
const int inf = 2e9 + 9;
const int N = 2e5 + 6;
int n, m, k;
vector<pair<int, int> > v[N];
ll hsh[N];
ll subset[10][10];
ll superset, curset, res;
void genPerm(int x) {
  if (x == k) {
    if (superset == curset) res++;
    return;
  }
  for (int i = 0; i <= x; i++) {
    auto temp = curset;
    (curset += subset[x][i]) %= sp2;
    genPerm(x + 1);
    curset = temp;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int I = 1; I <= m; I++) {
    int A, B, C;
    cin >> A >> B >> C;
    v[A].emplace_back(C, B);
  };
  for (int i = 1; i <= n; i++) sort(v[i].begin(), v[i].end());
  hsh[0] = 1;
  for (int i = 1; i <= n; i++) hsh[i] = (hsh[i - 1] * 7) % sp2;
  for (int i = 1; i <= n; i++) superset = (hsh[i] + superset) % sp2;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < v[i].size(); j++)
      (subset[v[i].size() - 1][j] += hsh[v[i][j].second]) %= sp2;
  }
  genPerm(0);
  cout << res << "\n";
}
ll power(ll x, ll y) {
  ll res = 1;
  x %= MOD;
  while (y > 0) {
    if (y & 1) res = (res * x) % MOD;
    y = y >> 1, x = (x * x) % MOD;
  }
  return res;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
using ll = long long;
using ld = long double;
template <typename T>
using pxx = pair<T, T>;
ll power(ll x, ll y);
const ll sp1 = 1e12 + 39;
const ll sp2 = 1e15 + 1e5 + 11;
const ll INF = 1e18 + 18;
const ll MOD = 1e9 + 7;
const int inf = 2e9 + 9;
const int N = 2e5 + 6;
int n, m, k;
vector<pair<int, int> > v[N];
ll hsh[N];
ll subset[10][10];
ll superset, curset, res;
void genPerm(int x) {
  if (x == k) {
    if (superset == curset) res++;
    return;
  }
  for (int i = 0; i <= x; i++) {
    auto temp = curset;
    (curset += subset[x][i]) %= sp2;
    genPerm(x + 1);
    curset = temp;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (int I = 1; I <= m; I++) {
    int A, B, C;
    cin >> A >> B >> C;
    v[A].emplace_back(C, B);
  };
  for (int i = 1; i <= n; i++) sort(v[i].begin(), v[i].end());
  hsh[0] = 1;
  for (int i = 1; i <= n; i++) hsh[i] = (hsh[i - 1] * 7) % sp2;
  for (int i = 1; i <= n; i++) superset = (hsh[i] + superset) % sp2;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < v[i].size(); j++)
      (subset[v[i].size() - 1][j] += hsh[v[i][j].second]) %= sp2;
  }
  genPerm(0);
  cout << res << "\n";
}
ll power(ll x, ll y) {
  ll res = 1;
  x %= MOD;
  while (y > 0) {
    if (y & 1) res = (res * x) % MOD;
    y = y >> 1, x = (x * x) % MOD;
  }
  return res;
}

```