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;
void canhazfast() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
}
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T extgcd(T a, T b, T &x, T &y) {
  T x0 = 1, y0 = 0, x1 = 0, y1 = 1;
  while (b) {
    T q = a / b;
    a %= b;
    swap(a, b);
    x0 -= q * x1;
    swap(x0, x1);
    y0 -= q * y1;
    swap(y0, y1);
  }
  x = x0;
  y = y0;
  return a;
}
int ctz(unsigned x) { return __builtin_ctz(x); }
int ctzll(unsigned long long x) { return __builtin_ctzll(x); }
int clz(unsigned x) { return __builtin_clz(x); }
int clzll(unsigned long long x) { return __builtin_clzll(x); }
int popcnt(unsigned x) { return __builtin_popcount(x); }
int popcntll(unsigned long long x) { return __builtin_popcountll(x); }
int bsr(unsigned x) { return 31 ^ clz(x); }
int bsrll(unsigned long long x) { return 63 ^ clzll(x); }
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second < b.second;
}
const int ofs[10] = {0, 1, 3, 6, 10, 15, 21, 28, 36};
int cur[10];
long long bad[10][10];
vector<pair<int, int> > out[200016];
vector<int> in[200016];
int lookup(int s, int t) {
  for (int i = 0; i < (int)out[s].size(); ++i) {
    if (out[s][i].first == t) return i;
  }
  return -1;
}
int dfs(int n, int k, long long msk = 0, int sz = 0) {
  if (sz == k) return 1;
  int res = 0;
  for (int i = 0; i <= sz; ++i) {
    long long bit = 1LL << (ofs[sz] + i);
    if (msk & bit) continue;
    cur[sz] = i;
    res += dfs(n, k, msk | bad[sz][i], sz + 1);
  }
  return res;
}
int main() {
  canhazfast();
  int n, m, k;
  long long dis = 0;
  cin >> n >> m >> k;
  for (int i = 0; i < m; ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    out[u].emplace_back(v, w);
    in[v].push_back(u);
  }
  for (int i = 1; i <= n; ++i) {
    if (out[i].empty() || in[i].empty()) return cout << "0\n", 0;
  }
  for (int i = 1; i <= n; ++i) sort((out[i]).begin(), (out[i]).end(), cmp);
  for (int i = 1; i <= n; ++i) {
    long long msk = 0;
    for (int j : in[i]) {
      size_t sz = out[j].size() - 1;
      int pos = lookup(j, i);
      long long bit = 1LL << (ofs[sz] + pos);
      dis |= bit & msk;
      msk |= bit;
    }
    for (int sz = 0; sz < k; ++sz) {
      for (int pos = 0; pos <= sz; ++pos) {
        long long bit = 1LL << (ofs[sz] + pos);
        if (msk & bit) bad[sz][pos] |= msk;
      }
    }
  }
  cout << dfs(n, k, dis) << '\n';
  return 0;
}
