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>
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("avx2")
#pragma GCC optimize("03")
using namespace std;
vector<pair<int, int> > g[(long long)(2e5 + 5)];
set<pair<int, int> > ig[(long long)(2e5 + 5)];
map<pair<int, int>, int> fq;
int cant[(long long)(2e5 + 5)];
int ma[100][100];
int trans(pair<int, int> x) {
  return ((x.first) * (x.first - 1)) / 2 + x.second;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    u--, v--;
    g[u].push_back(pair<int, int>(w, v));
  }
  for (int i = 0; i < n; i++) {
    sort((g[i]).begin(), (g[i]).end());
    for (int j = 0; j < g[i].size(); j++) {
      int u = g[i][j].second;
      cant[u]++;
      ig[u].insert(pair<int, int>(g[i].size(), j));
    }
  }
  bool ok = 1;
  int sum = 0;
  for (int i = 0; i < n; i++) {
    if (!cant[i]) ok = 0;
    sum += ig[i].size();
    vector<pair<int, int> > ks;
    for (auto y : ig[i]) {
      fq[y]++;
      ks.push_back(y);
    }
    for (int j = 0; j < ks.size(); j++)
      for (int u = j + 1; u < ks.size(); u++)
        ma[trans(ks[j])][trans(ks[u])] = ma[trans(ks[u])][trans(ks[j])] = 1;
  }
  if (!ok) {
    cout << (0) << '\n';
    return 0;
  }
  long long first[k + 1];
  first[0] = 1;
  for (int i = 1; i <= k; i++) first[i] = first[i - 1] * i;
  long long res = 0;
  for (int i = 0; i < first[k]; i++) {
    vector<int> p;
    int ff = i;
    for (int j = 0; j < k; j++) {
      p.push_back(ff / (first[k] / first[j + 1]));
      ff %= (first[k] / first[j + 1]);
    }
    long long sum = 0;
    for (int j = 0; j < k; j++) sum += fq[pair<int, int>(j + 1, p[j])];
    bool ok = 1;
    for (int j = 0; j < k; j++)
      for (int u = j + 1; u < k; u++)
        ok &= (!ma[trans(pair<int, int>(j + 1, p[j]))]
                  [trans(pair<int, int>(u + 1, p[u]))]);
    if (ok && sum == n) {
      res++;
    }
  }
  cout << (res) << '\n';
  return 0;
}
