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.
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  T sig = 1;
  for (; !isdigit(c); c = getchar())
    if (c == '-') sig = -1;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  x *= sig;
}
bool f[10][10][10][10];
class Solution {
  vector<vector<pair<int, int>>> adj;
  vector<vector<int>> in;
  int ans = 0, n, m, k;
  void dfs(vector<int> &choose, int d) {
    if (d == k + 1) {
      bool ok = true;
      for (int i = 1; i <= k; ++i)
        for (int j = 1; j <= k; ++j)
          if (f[i][choose[i]][j][choose[j]]) ok = false;
      if (ok) ans++;
      return;
    }
    if (choose[d] == 0) {
      for (int i = 1; i <= d; ++i) {
        bool ok = true;
        for (int j = 1; j < d; ++j)
          if (f[j][choose[j]][d][i]) {
            ok = false;
            break;
          }
        if (!ok) continue;
        choose[d] = i;
        dfs(choose, d + 1);
        choose[d] = 0;
      }
    } else
      dfs(choose, d + 1);
  }

 public:
  void solve() {
    read(n), read(m), read(k);
    adj = vector<vector<pair<int, int>>>(n + 1);
    in = vector<vector<int>>(n + 1);
    for (int i = 0; i < m; ++i) {
      int u, v, w;
      read(u), read(v), read(w);
      adj[u].emplace_back(w, v);
      in[v].emplace_back(u);
    }
    for (int i = 1; i <= n; ++i) sort(adj[i].begin(), adj[i].end());
    vector<int> choose(k + 1);
    memset(f, 0, sizeof(f));
    for (int i = 1; i <= n; ++i) {
      map<pair<int, int>, int> s;
      for (int u : in[i]) {
        int k = 0;
        while (adj[u][k].second != i) k++;
        int t = adj[u].size();
        s[{t, k + 1}]++;
      }
      for (auto &[p, pc] : s) {
        if (pc > 1) {
          for (int i = 1; i <= k; ++i)
            for (int j = 1; j <= i; ++j)
              f[i][j][p.first][p.second] = true,
              f[p.first][p.second][i][j] = true;
          continue;
        }
        for (auto &[q, qc] : s)
          if (p != q) f[p.first][p.second][q.first][q.second] = true;
      }
    }
    for (int i = 1; i <= n; ++i) {
      if (in[i].empty()) {
        printf("0");
        return;
      }
      if (in[i].size() == 1) {
        int u = in[i][0];
        int k = 0;
        while (adj[u][k].second != i) k++;
        int t = adj[u].size();
        if (choose[t] != 0 && choose[t] != k + 1) {
          printf("0");
          return;
        }
        choose[t] = k + 1;
      }
    }
    dfs(choose, 1);
    printf("%d", ans);
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  Solution solution = Solution();
  solution.solve();
}
