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 maxp = 22;
const long double EPS = 1e-18;
const long long INF = 1e18;
const int MOD = 1e9 + 7;
const int N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
uniform_int_distribution<long long> distribution;
long long rnd(long long l, long long r) {
  return distribution(rng) % (r - l + 1) + l;
}
const int HS = 3;
int mod[HS];
struct custom_hash {
  int val[HS];
  custom_hash() { fill(val, val + HS, 0); }
  custom_hash(long long x) {
    for (int i = 0; i < HS; i++) val[i] = (x % mod[i] + mod[i]) % mod[i];
  }
  custom_hash operator+(custom_hash x) {
    custom_hash res;
    res.val[0] = (val[0] + x.val[0]) % mod[0];
    res.val[1] = (val[1] + x.val[1]) % mod[1];
    res.val[2] = (1LL * val[2] * x.val[2]) % mod[2];
    return res;
  }
  bool operator==(const custom_hash &x) const {
    for (int i = 0; i < HS; i++)
      if (val[i] != x.val[i]) return false;
    return true;
  }
};
custom_hash s, val[N], sets[10][10];
int ans = 0;
int k;
vector<int> curr;
void solve(int st) {
  if (st == k) {
    custom_hash tot;
    for (int i = 1; i <= k; i++) tot = tot + sets[i][curr[i - 1]];
    if (tot == s) ans++;
    return;
  }
  for (int i = 1; i <= st + 1; i++) {
    curr.push_back(i);
    solve(st + 1);
    curr.pop_back();
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m >> k;
  mod[0] = MOD;
  mod[1] = 998244353;
  mod[2] = rnd(1e8, 1e9);
  int i, j;
  vector<vector<pair<int, int> > > a(n + 1, vector<pair<int, int> >());
  for (i = 0; i < m; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    a[x].push_back({z, y});
  }
  for (i = 1; i <= n; i++) val[i] = custom_hash(rnd(1, MOD));
  for (i = 1; i <= n; i++) {
    s = s + val[i];
    sort((a[i]).begin(), (a[i]).end());
    for (j = 0; j < a[i].size(); j++)
      sets[(int)a[i].size()][j + 1] =
          sets[(int)a[i].size()][j + 1] + val[a[i][j].second];
  }
  solve(0);
  cout << ans;
}
