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;
template <typename Tp>
inline void read(Tp &x) {
  static char c;
  static bool neg;
  x = 0, c = getchar(), neg = false;
  for (; !isdigit(c); c = getchar()) {
    if (c == '-') {
      neg = true;
    }
  }
  for (; isdigit(c); c = getchar()) {
    x = x * 10 + c - '0';
  }
  if (neg) {
    x = -x;
  }
}
const int N = 2e5 + 5;
vector<pair<int, int> > G[N];
mt19937_64 mtr(chrono::system_clock::now().time_since_epoch().count());
int n, m, k, ans = 0;
unsigned long long w[N], tot = 0;
unsigned long long a[10][10];
void dfs(int dep, unsigned long long now) {
  if (dep == k + 1) {
    if (now == tot) {
      ++ans;
    }
    return;
  }
  for (int i = 0; i < dep; ++i) {
    dfs(dep + 1, now ^ a[dep][i]);
  }
}
int main() {
  read(n), read(m), read(k);
  for (int i = 1, u, v, w; i <= m; ++i) {
    read(u), read(v), read(w);
    G[u].emplace_back((pair<int, int>){w, v});
  }
  for (int i = 1; i <= n; ++i) {
    tot ^= w[i] = mtr();
    sort(G[i].begin(), G[i].end());
  }
  for (int i = 1; i <= n; ++i) {
    for (int d = (int)G[i].size(), j = 0; j < d; ++j) {
      a[d][j] ^= w[G[i][j].second];
    }
  }
  dfs(1, 0);
  printf("%d\n", ans);
}
