#include <bits/stdc++.h>
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
using namespace std;
const int MAXN = 2e5 + 100;
struct edge {
  int nodo, cost, pos;
};
vector<edge> L[MAXN];
vector<pair<int, int>> arr[MAXN];
int N, M, K;
bool mk[10][10];
vector<pair<int, int>> temp[10][10];
int res;
void fb(int pos) {
  if (pos == K + 1) {
    res++;
    return;
  }
  for (int i = 1; i <= pos; i++) {
    vector<bool> tp;
    for (auto x : temp[pos][i]) {
      if (!mk[x.first][x.second]) {
        tp.push_back(true);
        mk[x.first][x.second] = true;
      } else {
        tp.push_back(false);
      }
    }
    if (!mk[pos][i]) fb(pos + 1);
    for (int x = 0; x < temp[pos][i].size(); x++) {
      if (tp[x] == true)
        mk[temp[pos][i][x].first][temp[pos][i][x].second] = false;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M >> K;
  for (int i = 0; i < M; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    L[a].push_back({b, c, -1});
  }
  for (int i = 1; i <= N; i++) {
    vector<pair<int, int>> temp;
    for (auto v : L[i]) {
      temp.push_back({v.cost, v.nodo});
    }
    L[i].clear();
    sort(temp.begin(), temp.end());
    for (int x = 0; x < temp.size(); x++) {
      L[i].push_back({temp[x].second, temp[x].first, x + 1});
    }
  }
  for (int i = 1; i <= N; i++) {
    for (auto x : L[i]) {
      arr[x.nodo].push_back({L[i].size(), x.pos});
    }
  }
  set<pair<pair<int, int>, pair<int, int>>> st;
  for (int i = 1; i <= N; i++) {
    sort(arr[i].begin(), arr[i].end());
    vector<pair<int, int>> vtt;
    int cont = 0;
    false;
    for (int x = 0; x < arr[i].size(); x++) {
      if (vtt.size() == 0) {
        vtt.push_back(arr[i][x]);
        cont = 1;
      } else {
        if (cont < 2) {
          if (vtt.back() == arr[i][x]) {
            cont++;
            vtt.push_back(arr[i][x]);
          } else {
            cont = 1;
            vtt.push_back(arr[i][x]);
          }
        } else {
          if (vtt.back() == arr[i][x]) {
          } else {
            cont = 1;
            vtt.push_back(arr[i][x]);
          }
        }
      }
    }
    false;
    arr[i] = vtt;
    for (int x = 0; x < arr[i].size(); x++) {
      for (int y = x + 1; y < arr[i].size(); y++) {
        pair<int, int> p1 = arr[i][x];
        pair<int, int> p2 = arr[i][y];
        if (p2 < p1) swap(p1, p2);
        st.insert({p1, p2});
      }
    }
  }
  vector<pair<pair<int, int>, pair<int, int>>> vect;
  for (auto x : st) {
    vect.push_back(x);
  }
  false;
  for (auto x : vect) {
    temp[x.first.first][x.first.second].push_back(x.second);
  }
  fb(1);
  cout << res << '\n';
  return 0;
}

### 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>
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
using namespace std;
const int MAXN = 2e5 + 100;
struct edge {
  int nodo, cost, pos;
};
vector<edge> L[MAXN];
vector<pair<int, int>> arr[MAXN];
int N, M, K;
bool mk[10][10];
vector<pair<int, int>> temp[10][10];
int res;
void fb(int pos) {
  if (pos == K + 1) {
    res++;
    return;
  }
  for (int i = 1; i <= pos; i++) {
    vector<bool> tp;
    for (auto x : temp[pos][i]) {
      if (!mk[x.first][x.second]) {
        tp.push_back(true);
        mk[x.first][x.second] = true;
      } else {
        tp.push_back(false);
      }
    }
    if (!mk[pos][i]) fb(pos + 1);
    for (int x = 0; x < temp[pos][i].size(); x++) {
      if (tp[x] == true)
        mk[temp[pos][i][x].first][temp[pos][i][x].second] = false;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M >> K;
  for (int i = 0; i < M; i++) {
    int a, b, c;
    cin >> a >> b >> c;
    L[a].push_back({b, c, -1});
  }
  for (int i = 1; i <= N; i++) {
    vector<pair<int, int>> temp;
    for (auto v : L[i]) {
      temp.push_back({v.cost, v.nodo});
    }
    L[i].clear();
    sort(temp.begin(), temp.end());
    for (int x = 0; x < temp.size(); x++) {
      L[i].push_back({temp[x].second, temp[x].first, x + 1});
    }
  }
  for (int i = 1; i <= N; i++) {
    for (auto x : L[i]) {
      arr[x.nodo].push_back({L[i].size(), x.pos});
    }
  }
  set<pair<pair<int, int>, pair<int, int>>> st;
  for (int i = 1; i <= N; i++) {
    sort(arr[i].begin(), arr[i].end());
    vector<pair<int, int>> vtt;
    int cont = 0;
    false;
    for (int x = 0; x < arr[i].size(); x++) {
      if (vtt.size() == 0) {
        vtt.push_back(arr[i][x]);
        cont = 1;
      } else {
        if (cont < 2) {
          if (vtt.back() == arr[i][x]) {
            cont++;
            vtt.push_back(arr[i][x]);
          } else {
            cont = 1;
            vtt.push_back(arr[i][x]);
          }
        } else {
          if (vtt.back() == arr[i][x]) {
          } else {
            cont = 1;
            vtt.push_back(arr[i][x]);
          }
        }
      }
    }
    false;
    arr[i] = vtt;
    for (int x = 0; x < arr[i].size(); x++) {
      for (int y = x + 1; y < arr[i].size(); y++) {
        pair<int, int> p1 = arr[i][x];
        pair<int, int> p2 = arr[i][y];
        if (p2 < p1) swap(p1, p2);
        st.insert({p1, p2});
      }
    }
  }
  vector<pair<pair<int, int>, pair<int, int>>> vect;
  for (auto x : st) {
    vect.push_back(x);
  }
  false;
  for (auto x : vect) {
    temp[x.first.first][x.first.second].push_back(x.second);
  }
  fb(1);
  cout << res << '\n';
  return 0;
}

```