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 bufSize = 1e6;
inline char nc() {
  return getchar();
  static char buf[bufSize], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, bufSize, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline void read(char *s) {
  static char c;
  for (; !isalpha(c); c = nc())
    ;
  for (; isalpha(c); c = nc()) *s++ = c;
  *s = '\0';
}
template <typename T>
inline T read(T &r) {
  static char c;
  static int flag;
  flag = 1, r = 0;
  for (c = nc(); !isdigit(c); c = nc())
    if (c == '-') flag = -1;
  for (; isdigit(c); c = nc()) r = r * 10 + c - 48;
  return r *= flag;
}
const int maxn = 2e5 + 100;
int n, m, k;
int sum;
int a[maxn];
int U[maxn], V[maxn];
vector<int> E[maxn];
int val[10][10];
int c[10], ans;
void dfs(int x) {
  if (x == k + 1) {
    int tsum = 0;
    for (int i = 1; i <= k; ++i) tsum += val[i][c[i]];
    if (sum == tsum) ++ans;
    return;
  }
  for (int i = 1; i <= x; ++i) c[x] = i, dfs(x + 1);
}
int main() {
  srand(time(0));
  read(n), read(m), read(k);
  for (int i = 1; i <= n; ++i) sum += (a[i] = rand());
  for (int i = 1; i <= m; ++i) {
    int a, b, c;
    read(a), read(b), read(c);
    U[c] = a, V[c] = b;
  }
  for (int i = 1; i <= m; ++i) E[U[i]].push_back(V[i]);
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= (int)E[i].size(); ++j)
      val[E[i].size()][j] += a[E[i][j - 1]];
  dfs(1);
  printf("%d\n", ans);
  return 0;
}
