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 maxn = 200010, maxk = 10, mod = 5111111;
int n, m, k;
long long ans(0);
vector<bitset<maxn> > hsh;
vector<unsigned long long> hshw;
vector<int> hshc;
vector<pair<int, int> > ver[maxn];
bitset<maxn> b[maxk][maxk];
bitset<maxn> now, U, E;
unsigned long long ghs(bitset<maxn> x) {
  unsigned long long d(0);
  for (register int i = 1; i <= n; ++i)
    if (x[i])
      d = d + ((unsigned long long)1 << i);
    else
      d = d + ((unsigned long long)3 << i);
  return d;
}
void dfs1(int x) {
  if (x <= 0) {
    unsigned long long d = ghs(now);
    int flag(0);
    for (register int i = 0; i < hsh.size(); ++i)
      if (d == hshw[i]) {
        if (now == hsh[i]) ++hshc[i];
        flag = 1;
      }
    if (!flag) hsh.push_back(now), hshw.push_back(d), hshc.push_back(1);
    return;
  }
  for (register int i = 1; i <= x; ++i) {
    if ((now & (b[x][i])) != E) continue;
    now |= b[x][i];
    dfs1(x - 2);
    now ^= b[x][i];
  }
}
void dfs2(int x) {
  if (x <= 0) {
    bitset<maxn> rec = U ^ now;
    unsigned long long d = ghs(rec);
    for (register int i = 0; i < hsh.size(); ++i)
      if (hshw[i] == d) {
        if (rec == hsh[i]) {
          ans = ans + hshc[i];
          break;
        }
      }
    return;
  }
  for (register int i = 1; i <= x; ++i) {
    if ((now & b[x][i]) != E) continue;
    now |= b[x][i];
    dfs2(x - 2);
    now ^= b[x][i];
  }
}
int main() {
  cin >> n >> m >> k;
  for (register int i = 1; i <= n; ++i) U[i] = 1;
  for (register int i = 1; i <= m; ++i) {
    int x, y, w;
    scanf("%d%d%d", &x, &y, &w);
    ver[x].push_back(make_pair(w, y));
  }
  for (register int i = 1; i <= n; ++i) {
    sort(ver[i].begin(), ver[i].end());
    for (register int j = 0; j < ver[i].size(); ++j)
      b[ver[i].size()][j + 1][ver[i][j].second] = 1;
  }
  dfs1(k);
  dfs2(k - 1);
  cout << ans;
  return 0;
}
