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;
inline int read() {
  int first = 0, f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar()) {
    if (ch == '-') f = -1;
  }
  for (; isdigit(ch); ch = getchar()) {
    first = first * 10 + ch - 48;
  }
  return first * f;
}
const int mxN = 2e5;
int n, m, p;
vector<pair<int, int> > adj[mxN + 3];
int oud[mxN + 3];
bitset<mxN + 3> f[13][13];
vector<int> vec[13];
bool con[13][13][13][13];
bool ban[13][13];
int seq[mxN + 3], ans;
void dfs1(int u) {
  if (u == p + 1) {
    bool ok = true;
    for (int i = 1; i <= p; i++)
      for (int j = i + 1; j <= p; j++) {
        if (con[i][seq[i]][j][seq[j]]) {
          ok = false;
          break;
        }
      }
    if (ok) {
      ans++;
    }
    return;
  }
  for (int i = 1; i <= u; i++)
    if (!ban[u][i]) {
      seq[u] = i;
      dfs1(u + 1);
    }
}
int main() {
  n = read(), m = read(), p = read();
  for (int i = 1; i <= m; i++) {
    int u = read(), v = read(), w = read();
    adj[u].push_back(make_pair(w, v));
    oud[u]++;
  }
  for (int u = 1; u <= n; u++) sort(adj[u].begin(), adj[u].end());
  for (int u = 1; u <= n; u++) vec[oud[u]].push_back(u);
  for (int i = 1; i <= p; i++)
    for (int j = 1; j <= i; j++) {
      for (int o = 0; o < vec[i].size(); o++) {
        int u = vec[i][o];
        if (f[i][j].test(adj[u][j - 1].second)) {
          ban[i][j] = true;
          break;
        }
        f[i][j].set(adj[u][j - 1].second);
      }
    }
  for (int i1 = 1; i1 <= p; i1++)
    for (int j1 = 1; j1 <= i1; j1++)
      for (int i2 = i1 + 1; i2 <= p; i2++)
        for (int j2 = 1; j2 <= i2; j2++) {
          if ((f[i1][j1] & f[i2][j2]).any()) {
            con[i1][j1][i2][j2] = true;
          }
        }
  dfs1(1);
  printf("%d\n", ans);
  return 0;
}
