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;
const int N = 2e5 + 5;
int n, m, k;
int head[N];
vector<pair<int, int>> g[N];
vector<int> f[10][10];
bool cross[10][10][10][10];
bool flag[10][10];
int ans;
void dfs(int x, int cnt = 0) {
  static int res[N];
  if (x > k) {
    if (cnt == n) {
      ans++;
    }
    return;
  }
  for (int i = 1; i <= x; ++i) {
    if (flag[x][i]) continue;
    bool flag = true;
    for (int j = 1; j < x; ++j)
      if (cross[x][i][j][res[j]]) {
        flag = false;
      }
    if (flag) {
      res[x] = i;
      dfs(x + 1, cnt + f[x][i].size());
    }
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 1, x, y, w; i <= m; ++i) {
    scanf("%d%d%d", &x, &y, &w);
    g[x].push_back({w, y});
  }
  for (int i = 1; i <= n; ++i) sort(g[i].begin(), g[i].end());
  for (int i = 1; i <= n; ++i) {
    int x = g[i].size();
    for (int j = 0; j < x; ++j) {
      f[x][j + 1].push_back(g[i][j].second);
    }
  }
  for (int i = 1; i <= k; ++i)
    for (int j = 1; j <= i; ++j) {
      static int tot;
      static int vis[N];
      ++tot;
      for (int t = 0; t < f[i][j].size(); ++t) {
        if (vis[f[i][j][t]] == tot) flag[i][j] = true;
        vis[f[i][j][t]] = tot;
      }
      if (flag[i][j]) continue;
      for (int i1 = 1; i1 < i; ++i1)
        for (int j1 = 1; j1 <= i1; ++j1) {
          for (int t = 0; t < f[i1][j1].size(); ++t)
            if (vis[f[i1][j1][t]] == tot) {
              cross[i][j][i1][j1] = true;
              break;
            }
        }
    }
  dfs(1);
  cout << ans << endl;
}
