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 = 200000, M = 200000, K = 9;
int oo[N], oi[N], *oho[N], *ohi[N];
int ww[M], aa[M], bb[M];
bool bad[K][K][K][K];
int aa_[K];
int dd[K][K];
int cnt;
void init(int n) {
  for (int i = 0; i < n; i++) {
    oho[i] = (int *)malloc(2 * sizeof *oho[i]);
    ohi[i] = (int *)malloc(2 * sizeof *ohi[i]);
  }
}
void link(int *oo, int **oh, int i, int h) {
  int &o = oo[i];
  if (o >= 2 && (o & o - 1) == 0)
    oh[i] = (int *)realloc(oh[i], (o << 1) * sizeof *oh[i]);
  oh[i][o++] = h;
}
void count(int k, int b) {
  if (b == k) {
    cnt++;
    return;
  }
  for (int a = 0; a <= b; a++) {
    if (bad[b][a][b][a]) continue;
    for (int b_ = 0; b_ < b; b_++) {
      int a_ = aa_[b_];
      if (bad[b_][a_][b][a]) goto end;
    }
    aa_[b] = a;
    count(k, b + 1);
  end:;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m, k;
  cin >> n >> m >> k;
  init(n);
  for (int h = 0; h < m; h++) {
    int i, j, w;
    cin >> i >> j >> w, i--, j--;
    ww[h] = w;
    link(oo, oho, i, h);
    link(oi, ohi, j, h);
  }
  for (int i = 0; i < n; i++) {
    sort(oho[i], oho[i] + oo[i],
         [](int h1, int h2) { return ww[h1] < ww[h2]; });
    for (int o = 0; o < oo[i]; o++) {
      int h = oho[i][o];
      aa[h] = o;
      bb[h] = oo[i] - 1;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int b = 0; b < k; b++)
      for (int a = 0; a <= b; a++) dd[b][a] = 0;
    for (int o = 0; o < oi[i]; o++) {
      int h = ohi[i][o], a = aa[h], b = bb[h];
      dd[b][a]++;
    }
    for (int b1 = 0; b1 < k; b1++)
      for (int a1 = 0; a1 <= b1; a1++) {
        if (dd[b1][a1] != 1) {
          if (dd[b1][a1] > 1) bad[b1][a1][b1][a1] = true;
          continue;
        }
        for (int b2 = b1 + 1; b2 < k; b2++)
          for (int a2 = 0; a2 <= b2; a2++) {
            if (dd[b2][a2] == 0) continue;
            bad[b1][a1][b2][a2] = true;
          }
      }
  }
  count(k, 0);
  cout << cnt << '\n';
  return 0;
}
