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 = 200000;
const int MAXK = 9;
const int MAXH = 89;
int n, m, k;
int ans = 0;
int c[MAXK + 5];
vector<int> in[MAXN + 5];
bool bad[MAXH + 5][MAXH + 5];
vector<pair<int, int> > G[MAXN + 5];
void gen(int i) {
  if (i == k + 1) {
    ans++;
    return;
  }
  for (int j = 0; j < i; ++j) {
    int hash = i * 9 + j;
    if (bad[hash][hash]) continue;
    bool flag = true;
    for (int k = 1; k < i; ++k) {
      if (bad[c[k]][hash]) {
        flag = false;
        break;
      }
    }
    if (!flag) continue;
    c[i] = hash;
    gen(i + 1);
  }
}
int main() {
  ios ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    G[u].push_back(make_pair(w, v));
  }
  for (int i = 1; i <= n; ++i) {
    sort(G[i].begin(), G[i].end());
    int sz = G[i].size();
    for (int j = 0; j < sz; ++j) {
      in[G[i][j].second].push_back(sz * 9 + j);
    }
  }
  for (int i = 1; i <= n; ++i) {
    int sz = in[i].size();
    if (sz == 0) {
      cout << 0 << '\n';
      return 0;
    }
    vector<int> v = in[i];
    in[i].clear();
    in[i].push_back(v[0]);
    for (int j = 1; j < sz; ++j) {
      if (v[j] == v[j - 1])
        bad[v[j]][v[j]] = true;
      else
        in[i].push_back(v[j]);
    }
    sz = in[i].size();
    for (int j = 0; j < sz; ++j) {
      for (int k = j + 1; k < sz; ++k) {
        bad[in[i][j]][in[i][k]] = bad[in[i][k]][in[i][j]] = true;
      }
    }
  }
  gen(1);
  cout << ans << '\n';
  return 0;
}
