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 MaxN(200003);
struct edge_t {
  int v, w;
  edge_t(int _v = 0, int _w = 0) : v(_v), w(_w) {}
  inline bool operator<(const edge_t& T) const { return w < T.w; }
} E[MaxN][11];
int d[MaxN], n, m, k;
bitset<MaxN> S[11][11];
int res[11], C[11][11], ans;
bool F[11][11][11][11];
void dfs(int i) {
  if (i > k) {
    bool Flag = true;
    int cnt = 0;
    for (int i = 1; i <= k && Flag; i++) {
      cnt += C[i][res[i]];
      for (int j = i + 1; j <= k; j++)
        if (F[i][res[i]][j][res[j]]) {
          Flag = false;
          break;
        }
    }
    if (Flag && cnt == n) ++ans;
    return;
  }
  for (int j = 1; j <= i; j++) res[i] = j, dfs(i + 1);
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int u, v, w; m--;) {
    scanf("%d%d%d", &u, &v, &w);
    E[u][++d[u]] = edge_t(v, w);
  }
  for (int i = 1; i <= n; i++) {
    sort(E[i] + 1, E[i] + 1 + d[i]);
    for (int j = 1; j <= d[i]; j++) S[d[i]][j][E[i][j].v - 1] = 1;
  }
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= i; j++) {
      C[i][j] = S[i][j].count();
      for (int x = j + 1; x <= k; x++)
        for (int y = 1; y <= x; y++)
          if ((S[i][j] & S[x][y]).count()) F[i][j][x][y] = 1;
    }
  dfs(1);
  printf("%d\n", ans);
  return 0;
}
