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;
namespace IO {
template <class T>
inline void read(T& ret) {
  ret = 0;
  int f = 1;
  char c = getchar();
  while (c < 48 || c > 57) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= 48 && c <= 57) ret = ret * 10 + c - 48, c = getchar();
  ret *= f;
}
template <class T>
inline void prin(const T& ret) {
  if (!ret) {
    putchar(48);
    return;
  }
  static T o;
  o = ret;
  if (o < 0) {
    putchar('-');
    o = -o;
  }
  static int op[25], tt;
  tt = 0;
  while (o) {
    op[++tt] = o % 10;
    o /= 10;
  }
  for (int i = tt; i; --i) putchar(op[i] + 48);
}
template <class T>
inline void pri_(const T& ret) {
  prin(ret);
  putchar(' ');
}
template <class T>
inline void priE(const T& ret) {
  prin(ret);
  putchar('\n');
}
}  // namespace IO
using namespace IO;
int n, m, k;
vector<pair<int, int> > e[200005];
unsigned long long G1[15][15], G2[15][15], G3[15][15];
int val1[200005], val2[200005], val3[200005], val4[200005];
int a[15];
int ans = 0;
int tmp1, tmp2;
void dfs(int u) {
  if (u == k + 1) {
    if (tmp2 == 0) ans++;
    return;
  }
  for (int i = 1; i <= u; i++) {
    a[u] = i;
    tmp2 ^= G2[u][a[u]];
    dfs(u + 1);
    tmp2 ^= G2[u][a[u]];
  }
}
int main() {
  srand(time(0));
  read(n);
  read(m);
  read(k);
  for (int i = 1; i <= m; i++) {
    int u, v, w;
    read(u);
    read(v);
    read(w);
    e[u].push_back(make_pair(w, v));
  }
  for (int i = 1; i <= n; i++) {
    val1[i] = (unsigned long long)rand() + 1;
    val2[i] = (unsigned long long)rand() * rand() + 1;
    val3[i] = (unsigned long long)rand() * rand() * rand() + 1;
    val4[i] = (unsigned long long)rand() * rand() * rand() * rand() + 1;
  }
  for (int i = 1; i <= n; i++) {
    if (!e[i].size()) {
      puts("0");
      return 0;
    }
    sort(e[i].begin(), e[i].end());
    int cnt = 0;
    for (auto j : e[i]) {
      ++cnt;
      G2[e[i].size()][cnt] ^= (val3[j.second] ^ val3[i]);
    }
  }
  dfs(1);
  cout << ans << endl;
}
