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 N = 3e5 + 5, mod = 1e9 + 7;
template <class o>
inline void qr(o &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + (c ^ 48);
    c = getchar();
  }
  x *= f;
}
template <class o>
void qw(o x) {
  if (x / 10) qw(x / 10);
  putchar(x % 10 + 48);
}
template <class o>
void pr1(o x) {
  if (x < 0) putchar('-'), x = -x;
  qw(x);
  putchar(' ');
}
template <class o>
void pr2(o x) {
  if (x < 0) putchar('-'), x = -x;
  qw(x);
  puts("");
}
inline int ksm(int a, int b = mod - 2) {
  int ans = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) ans = 1ll * ans * a % mod;
  return ans;
}
vector<pair<int, int> > e[N];
int H[11][11], val[N], tmp = 1, ans, n, m, k;
void dfs(int i, int s) {
  if (i == k + 1) {
    if (s == tmp) ++ans;
    return;
  }
  for (int j = 1; j <= i; ++j) dfs(i + 1, 1ll * s * H[i][j] % mod);
}
void solve() {
  qr(n), qr(m), qr(k);
  for (int i = 1; i <= m; ++i) {
    int x, y, w;
    qr(x), qr(y), qr(w);
    e[x].push_back(make_pair(w, y));
  }
  for (int i = 1; i <= n; ++i) sort(e[i].begin(), e[i].end());
  for (int i = 1; i <= k; ++i)
    for (int j = 1; j <= i; ++j) H[i][j] = 1;
  for (int i = 1; i <= n; ++i)
    val[i] = (1ll * ((i << 3) + (i >> 2)) * 173 + 23338297) % 932456477,
    tmp = 1ll * tmp * val[i] % mod;
  for (int i = 1; i <= n; ++i) {
    int s = e[i].size();
    for (int j = 0; j < s; ++j)
      H[s][j + 1] = 1ll * H[s][j + 1] * val[e[i][j].second] % mod;
  }
  dfs(1, 1);
  pr2(ans);
}
int main() {
  solve();
  return 0;
}
