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 inf = (int)1e9;
const int mod = inf + 7;
const double eps = 1e-9;
const double pi = acos(-1.0);
int bad[10][10];
int n, m, k;
vector<pair<int, int> > g[200200], gr[200200];
map<pair<int, int>, int> id;
map<pair<pair<int, int>, pair<int, int> >, int> can;
bool cmp(pair<int, int> a, pair<int, int> b) {
  return make_pair(a.second, a.first) < make_pair(b.second, b.first);
}
int choose[111];
void rec(int ind, int &ans) {
  if (ind > k) {
    ans++;
    return;
  }
  for (int i = 0; i < ind; i++) {
    if (bad[ind][i]) continue;
    int good = true;
    for (int r = 1; r < ind; r++) {
      int c = choose[r];
      if (can[make_pair(make_pair(r, c), make_pair(ind, i))]) good = false;
    }
    if (!good) continue;
    choose[ind] = i;
    rec(ind + 1, ans);
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0, x, y, z; i < m; i++) {
    scanf("%d%d%d", &x, &y, &z);
    x--;
    y--;
    g[x].push_back(make_pair(y, z));
    gr[y].push_back(make_pair(x, z));
  }
  for (int i = 0; i < n; i++) {
    sort(g[i].begin(), g[i].end(), cmp);
    for (int j = 0; j < g[i].size(); j++) {
      id[make_pair(i, g[i][j].first)] = j;
    }
  }
  for (int v = 0; v < n; v++) {
    map<pair<int, int>, int> cnt;
    for (__typeof(gr[v].begin()) it = gr[v].begin(); it != gr[v].end(); it++) {
      int u = (*it).first;
      int p = id[make_pair(u, v)];
      int q = g[u].size();
      cnt[make_pair(p, q)]++;
    }
    vector<pair<int, int> > tmp;
    for (__typeof(cnt.begin()) it = cnt.begin(); it != cnt.end(); it++) {
      pair<int, int> p = it->first;
      int c = it->second;
      if (c >= 2) bad[p.second][p.first] = 1;
      tmp.push_back(make_pair(p.second, p.first));
    }
    for (int i = 0; i < tmp.size(); i++) {
      for (int j = i + 1; j < tmp.size(); j++) {
        can[make_pair(tmp[i], tmp[j])] = can[make_pair(tmp[j], tmp[i])] = 1;
      }
    }
  }
  int ans = 0;
  rec(1, ans);
  cout << ans << endl;
  return 0;
}
