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 MAXN = 2e5 + 19;
int n, m, k;
int id[10][10], pj;
std::vector<std::pair<int, int> > to[MAXN], has[MAXN];
bool disable[10][10];
long long hash[10][10], st, ans, ban;
void dfs(int node) {
  if (node > k) {
    ++ans;
    return;
  }
  for (int i = 1; i <= node; ++i) {
    if (!disable[node][i] && !(st & hash[node][i]) &&
        !(ban & (1ll << id[node][i]))) {
      st ^= 1ll << id[node][i];
      long long tmp = ban;
      ban |= hash[node][i];
      dfs(node + 1);
      st ^= 1ll << id[node][i];
      ban = tmp;
    }
  }
}
int main() {
  std::scanf("%d%d%d", &n, &m, &k);
  for (int i = 1; i <= k; ++i)
    for (int j = 1; j <= i; ++j) id[i][j] = pj++;
  for (int i = 1; i <= m; ++i) {
    int u, v, w;
    std::scanf("%d%d%d", &u, &v, &w);
    to[u].push_back({w, v});
  }
  for (int i = 1; i <= n; ++i) {
    std::sort(to[i].begin(), to[i].end());
    for (int j = 1; j <= to[i].size(); ++j)
      has[to[i][j - 1].second].push_back({to[i].size(), j});
  }
  for (int i = 1; i <= n; ++i) {
    std::sort(has[i].begin(), has[i].end());
    for (int j = 1; j < has[i].size(); ++j)
      if (has[i][j - 1] == has[i][j])
        disable[has[i][j].first][has[i][j].second] = 1;
    has[i].resize(std::unique(has[i].begin(), has[i].end()) - has[i].begin());
    long long tmp = 0;
    for (int j = 0; j < has[i].size(); ++j)
      tmp ^= 1ll << id[has[i][j].first][has[i][j].second];
    for (int j = 0; j < has[i].size(); ++j)
      hash[has[i][j].first][has[i][j].second] |= tmp;
  }
  dfs(1);
  std::printf("%lld\n", ans);
}
