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;
constexpr int mxn = 2e5 + 5;
int n, m, k, cnt[10][10];
bool bad[10][10], bad_pairs[10][10][10][10];
vector<pair<int, int>> out[mxn], in[mxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    out[u].emplace_back(w, v);
  }
  for (int i = 1; i <= n; i++) {
    sort(out[i].begin(), out[i].end());
    int degree = out[i].size();
    for (int j = 0; j < degree; j++)
      in[out[i][j].second].emplace_back(degree, j + 1);
  }
  for (int u = 1; u <= n; u++) {
    int c[10][10] = {};
    for (const auto& p : in[u]) c[p.first][p.second]++;
    for (int i = 1; i <= k; i++) {
      for (int j = 1; j <= k; j++) {
        if (c[i][j] > 1) {
          bad[i][j] = true;
        } else if (c[i][j] == 1) {
          cnt[i][j]++;
          for (int x = i + 1; x <= k; x++)
            for (int y = 1; y <= x; y++)
              if (c[x][y] == 1) bad_pairs[i][j][x][y] = true;
        }
      }
    }
  }
  int all = 1, ans = 0;
  for (int i = 2; i <= k; i++) all *= i;
  for (int idx = 0; idx < all; idx++) {
    int sum = 0, cur_idx = idx;
    bool ok = true, became_bad[10][10] = {};
    for (int i = 1, j; i <= k; i++) {
      j = (cur_idx % i) + 1;
      cur_idx /= i;
      if (bad[i][j] || became_bad[i][j]) {
        ok = false;
        break;
      }
      sum += cnt[i][j];
      for (int x = i + 1; x <= k; x++)
        for (int y = 1; y <= x; y++) became_bad[x][y] |= bad_pairs[i][j][x][y];
    }
    if (ok && sum == n) ans++;
  }
  cout << ans << '\n';
  return 0;
}
