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;
template <typename T>
inline void read(T &n) {
  T w = 1;
  n = 0;
  char ch = getchar();
  while (!isdigit(ch) && ch != EOF) {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (isdigit(ch) && ch != EOF) {
    n = (n << 1) + (n << 3) + (ch ^ 48);
    ch = getchar();
  }
  n *= w;
}
template <typename T>
inline void write(T x) {
  unsigned long long y = 0;
  int l = 0;
  if (x < 0) {
    x = -x;
    putchar('-');
  }
  if (!x) {
    putchar(48);
    return;
  }
  while (x) {
    y = y * 10 + x % 10;
    x /= 10;
    l++;
  }
  while (l) {
    putchar(y % 10 + 48);
    y /= 10;
    l--;
  }
}
template <typename T>
inline void writeln(T x) {
  write(x);
  puts("");
}
template <typename T>
inline void writes(T x) {
  write(x);
  putchar(' ');
}
template <typename T>
inline void checkmax(T &a, T b) {
  a = a > b ? a : b;
}
template <typename T>
inline void checkmin(T &a, T b) {
  a = a < b ? a : b;
}
const int N = 4e5 + 100, K = 15;
struct wzy {
  int to, val;
} h[N][K];
int s[K], cnt[N], ans;
unsigned long long hh[N], nh[K][K];
inline bool cmp(wzy x, wzy y) { return x.val < y.val; }
unsigned long long no = 0, zz = 0;
inline void dfs(int x, int k) {
  if (x > k) {
    ans += (no == zz);
    return;
  }
  for (int i = (1); i <= (x); i++) {
    no ^= nh[x][i];
    dfs(x + 1, k);
    no ^= nh[x][i];
  }
}
int main() {
  int n, m, k;
  read(n);
  read(m);
  read(k);
  srand(time(0));
  for (int i = (1); i <= (n); i++) {
    hh[i] = 1ULL * rand() * rand() * rand();
    zz ^= hh[i];
  }
  for (int i = (1); i <= (m); i++) {
    int x, y, w;
    read(x);
    read(y);
    read(w);
    cnt[x]++;
    h[x][cnt[x]].to = y;
    h[x][cnt[x]].val = w;
  }
  for (int i = (1); i <= (n); i++) sort(h[i] + 1, h[i] + cnt[i] + 1, cmp);
  for (int i = (1); i <= (n); i++)
    for (int j = (1); j <= (cnt[i]); j++) {
      nh[cnt[i]][j] ^= hh[h[i][j].to];
    }
  dfs(1, k);
  writeln(ans);
  return 0;
}
