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 <class T>
inline void read(T& a) {
  char c = getchar();
  int f = 1;
  a = 0;
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') a = (a << 1) + (a << 3) + c - 48, c = getchar();
  a *= f;
}
template <class T>
void write(T a) {
  if (a < 0) putchar('-'), a = -a;
  if (a > 9) write(a / 10);
  putchar(a % 10 + 48);
}
const int o = 2e5 + 10, MOD = 1e9 + 7;
int n, m, k, U, V, W, h[o], ans, lgl, f[10][10], l;
vector<pair<int, int> > e[o];
void dfs(int nw, int len) {
  if (nw > k) {
    if (len == lgl) ++ans;
    return;
  }
  for (int i = 1; i <= nw; ++i) dfs(nw + 1, (len + f[nw][i]) % MOD);
}
int main() {
  read(n);
  read(m);
  read(k);
  for (int i = 1; i <= m; ++i)
    read(U), read(V), read(W), e[U].push_back(make_pair(W, V));
  lgl = h[1] = 1;
  for (int i = 2; i <= n; ++i) lgl = (lgl + (h[i] = h[i - 1] * 2 % MOD)) % MOD;
  for (int i = 1; i <= n; ++i)
    if (l = e[i].size()) {
      sort(e[i].begin(), e[i].end());
      for (int j = 0; j < l; ++j)
        f[l][j + 1] = (f[l][j + 1] + h[e[i][j].second]) % MOD;
    };
  dfs(1, 0);
  write(ans);
  return 0;
}
