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>
const int MAX_N = 200000;
const int MAX_M = 200000;
struct Edge {
  int u, v, w;
  int restr;
} edges[MAX_M];
int get_restr(int cid, int cval) { return (cid - 1) * 9 + cval; }
std::vector<int> graphIn[1 + MAX_N], graphOut[1 + MAX_N];
bool cmp(int a, int b) { return edges[a].w < edges[b].w; }
bool table[9 * 9][9 * 9];
int fr[9 * 9];
int res = 0;
int restr[9];
void bkt(int i, int k) {
  if (i == k)
    res++;
  else {
    for (int t = i * 9; t < i * 9 + i + 1; ++t) {
      restr[i] = t;
      bool ok = true;
      for (int j = 0; j <= i; ++j)
        if (table[restr[i]][restr[j]]) ok = false;
      if (ok) bkt(i + 1, k);
    }
  }
}
int main() {
  int n, m, k;
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0; i < m; ++i) {
    scanf("%d%d%d", &edges[i].u, &edges[i].v, &edges[i].w);
    graphIn[edges[i].v].push_back(i);
    graphOut[edges[i].u].push_back(i);
  }
  for (int i = 1; i <= n; ++i) {
    std::sort(graphOut[i].begin(), graphOut[i].end(), cmp);
    int val = 0;
    for (auto it : graphOut[i]) {
      edges[it].restr = get_restr(graphOut[i].size(), val);
      ++val;
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (int i = 0; i < 9 * 9; ++i) fr[i] = 0;
    for (auto it : graphIn[i]) fr[edges[it].restr]++;
    for (auto it : graphIn[i]) {
      for (int i = 0; i < 9 * 9; ++i) {
        if (edges[it].restr == i && fr[i] > 1)
          table[i][i] = true;
        else if (edges[it].restr != i && fr[i] > 0)
          table[edges[it].restr][i] = true;
      }
    }
  }
  bkt(0, k);
  printf("%d", res);
  return 0;
}
