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;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  vector<vector<pair<int, int>>> g(n);
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    --u, --v;
    g[u].push_back({w, v});
  }
  for (int i = 0; i < n; ++i) {
    sort(g[i].begin(), g[i].end());
  }
  vector<unsigned long long> h(n);
  for (int i = 0; i < n; ++i) {
    h[i] = rnd();
  }
  unsigned long long all = 0;
  for (int i = 0; i < n; ++i) all += h[i];
  vector<vector<unsigned long long>> sum(k + 1,
                                         vector<unsigned long long>(k + 1));
  for (int d = 1; d <= k; ++d) {
    for (int x = 1; x <= d; ++x) {
      for (int i = 0; i < n; ++i) {
        if ((int)g[i].size() == d) {
          sum[d][x] += h[g[i][x - 1].second];
        }
      }
    }
  }
  int ans = 0;
  function<void(int, unsigned long long)> rec = [&](int pos,
                                                    unsigned long long sumh) {
    if (pos == k + 1) {
      if (sumh == all) ++ans;
    } else {
      for (int x = 1; x <= pos; ++x) {
        rec(pos + 1, sumh + sum[pos][x]);
      }
    }
  };
  rec(1, 0);
  cout << ans << '\n';
}
