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;
int n, m, k, ans = 0, t1 = 0, t2 = 1;
int x[200002], y[200002], w[200002], id[200002];
int Hs1[9][9], Hs2[9][9];
vector<int> vec[200002];
template <class T>
void read(T &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  x = f ? -x : x;
  return;
}
inline bool cmp(int a, int b) { return w[a] < w[b] ? 1 : 0; }
inline void dfs(int x, int y1, int y2) {
  if (x > k) {
    ans += (y1 == t1 && y2 == t2);
    return;
  }
  for (int i = 0; i < x; ++i) dfs(x + 1, y1 + Hs1[x][i], y2 * Hs2[x][i]);
}
int main() {
  read(n), read(m), read(k);
  for (int i = 1; i <= m; ++i) read(x[i]), read(y[i]), read(w[i]), id[i] = i;
  sort(id + 1, id + m + 1, cmp);
  for (int i = 1; i <= m; ++i) vec[x[id[i]]].push_back(y[id[i]]);
  for (int i = 1; i <= k; ++i)
    for (int j = 0; j < i; ++j) Hs1[i][j] = 0, Hs2[i][j] = 1;
  for (int i = 1; i <= n; ++i) {
    t1 += i, t2 *= i;
    for (int j = 0; j < vec[i].size(); ++j)
      Hs1[vec[i].size()][j] += vec[i][j], Hs2[vec[i].size()][j] *= vec[i][j];
  }
  dfs(1, 0, 1), printf("%d", ans);
  return 0;
}
