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 pii = std::pair<int, int>;
using pll = std::pair<long long, long long>;
const int maxn = 2e5 + 100;
using bs = std::bitset<maxn>;
std::vector<pii> out[maxn];
std::vector<pii> in[maxn];
int bad[10][10];
int bad2[10][10][10][10];
std::vector<pii> active;
int generate(int i, int k) {
  if (i == k + 1) return 1;
  int result = 0;
  for (int v = 1; v <= i; v++) {
    if (bad[i][v]) continue;
    bool ok = true;
    for (auto& a : active) {
      if (bad2[a.first][a.second][i][v]) ok = false;
    }
    if (ok) {
      active.push_back({i, v});
      result += generate(i + 1, k);
      active.pop_back();
    }
  }
  return result;
}
int main() {
  std::ios::sync_with_stdio(false);
  int n, m, k;
  std::cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int x, y, w;
    std::cin >> x >> y >> w;
    out[x].push_back({w, y});
  }
  for (int i = 1; i <= n; i++) std::sort(out[i].begin(), out[i].end());
  std::map<int, std::vector<int>> by_deg;
  for (int i = 1; i <= n; i++) {
    by_deg[out[i].size()].push_back(i);
  }
  for (int i = 1; i <= k; i++) {
    for (int v = 1; v <= i; v++) {
      std::vector<int> outs;
      for (auto& x : by_deg[i]) {
        int y = out[x][v - 1].second;
        outs.push_back(y);
      }
      std::sort(outs.begin(), outs.end());
      outs.erase(std::unique(outs.begin(), outs.end()), outs.end());
      if (outs.size() != by_deg[i].size()) {
        bad[i][v] = 1;
        continue;
      }
      for (int y : outs) {
        for (auto& prev : in[y]) {
          bad2[prev.first][prev.second][i][v] = 1;
          bad2[i][v][prev.first][prev.second] = 1;
        }
        in[y].push_back({i, v});
      }
    }
  }
  std::cout << generate(1, k) << std::endl;
  return 0;
}
