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>
#pragma GCC optimize(2)
using namespace std;
inline long long read() {
  char c = getchar();
  long long x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
  return x;
}
long long res, n, m, k;
vector<pair<long long, long long> > e[200005];
vector<long long> v[13];
const long long b1 = 19491001;
unsigned long long pw[200005], sumn, sum[13][13], qwq;
void dfs(long long u) {
  if (u == k + 1) {
    res += (qwq == sumn);
    return;
  }
  for (register long long i = (0); i <= (u - 1); ++i) {
    qwq += sum[u][i];
    dfs(u + 1);
    qwq -= sum[u][i];
  }
}
signed main() {
  n = read(), m = read(), k = read();
  pw[0] = 1;
  for (register long long i = (1); i <= (n); ++i)
    pw[i] = pw[i - 1] * b1, sumn += pw[i];
  for (register long long i = (1); i <= (m); ++i) {
    long long u = read(), v = read(), w = read();
    e[u].push_back(make_pair(w, v));
  }
  for (register long long i = (1); i <= (n); ++i)
    sort(e[i].begin(), e[i].end()), v[e[i].size()].push_back(i);
  for (register long long i = (1); i <= (k); ++i)
    for (register long long j = (0); j <= (i - 1); ++j) {
      unsigned long long now = 0;
      for (long long p = 0; p < v[i].size(); ++p)
        now += pw[e[v[i][p]][j].second];
      sum[i][j] = now;
    }
  dfs(1);
  cout << res;
  return 0;
}
