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 MAXN = 2e5 + 10, K = 15;
vector<pair<int, int>> g[MAXN];
vector<int> vs[K][K];
int can[K][K][K][K], valid[K][K], pth[K], k;
bool check(vector<int> &a, vector<int> &b) {
  for (auto x : b) {
    int it = lower_bound(a.begin(), a.end(), x) - a.begin();
    if (it < int(a.size()) && a[it] == x) return 0;
  }
  return 1;
}
int go(int pos) {
  if (pos == k + 1) return 1;
  int ans = 0;
  for (int i = 0, ThxDem = pos; i < ThxDem; ++i) {
    int f = valid[pos][i];
    for (int j = 1, ThxDem = pos; j < ThxDem; ++j) {
      f = f && pth[j] != -1 && can[j][pth[j]][pos][i];
    }
    if (f) {
      pth[pos] = i;
      ans += go(pos + 1);
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m >> k;
  memset(pth, -1, sizeof(pth));
  for (int i = 0, ThxDem = m; i < ThxDem; ++i) {
    int x, y, w;
    cin >> x >> y >> w;
    x--;
    y--;
    g[x].push_back({w, y});
  }
  for (int i = 0, ThxDem = n; i < ThxDem; ++i) sort(g[i].begin(), g[i].end());
  for (int i = 0, ThxDem = n; i < ThxDem; ++i)
    for (int j = 0, ThxDem = int(g[i].size()); j < ThxDem; ++j)
      vs[int(g[i].size())][j].push_back(g[i][j].second);
  for (int i = 1, ThxDem = k + 1; i < ThxDem; ++i)
    for (int j = 0, ThxDem = i; j < ThxDem; ++j)
      sort(vs[i][j].begin(), vs[i][j].end());
  for (int i = 1, ThxDem = k + 1; i < ThxDem; ++i)
    for (int j = 0, ThxDem = i; j < ThxDem; ++j) {
      int pre = int(vs[i][j].size());
      vs[i][j].erase(unique(vs[i][j].begin(), vs[i][j].end()), vs[i][j].end());
      valid[i][j] = pre == int(vs[i][j].size());
    }
  for (int i = 1, ThxDem = k + 1; i < ThxDem; ++i)
    for (int j = i + 1, ThxDem = k + 1; j < ThxDem; ++j)
      for (int xi = 0, ThxDem = i; xi < ThxDem; ++xi)
        for (int xj = 0, ThxDem = j; xj < ThxDem; ++xj) {
          int asd = check(vs[i][xi], vs[j][xj]);
          can[i][xi][j][xj] = asd;
        }
  cout << go(1) << "\n";
}
