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 = 2e5 + 5;
int n, m, k, i, j, l, eu[N], ev[N], u, v, w, d1[N], d2[N], ans;
vector<int> e[N];
bool b[99][99], bb[99];
void dfs(int d) {
  if (d == k + 1) {
    ++ans;
    return;
  }
  for (int i = 1, v; i <= d; ++i)
    if (v = i + d * (d - 1) / 2, !b[v][v]) {
      for (j = 1; j < v; ++j)
        if (bb[j] && b[v][j]) break;
      if (j == v) bb[v] = 1, dfs(d + 1), bb[v] = 0;
    }
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (i = 1; i <= m; ++i)
    scanf("%d%d%d", &u, &v, &w), eu[w] = u, ev[w] = v, ++d1[u];
  for (i = 1; i <= m; ++i)
    u = eu[i], v = ev[i], e[v].push_back(++d2[u] + d1[u] * (d1[u] - 1) / 2);
  for (i = 1; i <= n; ++i) {
    sort(e[i].begin(), e[i].end());
    for (j = 0; j + 1 < e[i].size(); ++j)
      if (e[i][j] == e[i][j + 1]) b[e[i][j]][e[i][j]] = 1;
    e[i].erase(unique(e[i].begin(), e[i].end()), e[i].end());
    for (j = e[i].size() - 1; j >= 0; --j)
      for (int k = j - 1; k >= 0; --k) b[e[i][j]][e[i][k]] = 1;
  }
  dfs(1);
  printf("%d\n", ans);
}
