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 = 10;
int n, m, sk;
struct state {
  int x, y;
  state(int X = 0, int Y = 0) {
    x = X;
    y = Y;
  }
  bool operator<(const state &p) const {
    return y < p.y || y == p.y && x < p.x;
  }
};
vector<state> e[N + 9];
int deg[N + 9];
set<state> s[N + 9];
int b[M + 9][M + 9][M + 9][M + 9];
int vis[M + 9][M + 9];
int ans;
void Dfs_ans(int k) {
  if (k > sk) {
    ++ans;
    return;
  }
  for (int i = 1; i <= k; ++i)
    if (!vis[k][i]) {
      for (int j = 1; j <= sk; ++j)
        for (int t = 1; t <= sk; ++t)
          if (b[k][i][j][t]) ++vis[j][t];
      Dfs_ans(k + 1);
      for (int j = 1; j <= sk; ++j)
        for (int t = 1; t <= sk; ++t)
          if (b[k][i][j][t]) --vis[j][t];
    }
}
state now[N + 9];
int cn;
void work() {
  scanf("%d%d%d", &n, &m, &sk);
  for (int i = 1; i <= m; ++i) {
    int x, y, v;
    scanf("%d%d%d", &x, &y, &v);
    e[x].push_back(state(y, v));
    ++deg[x];
  }
  for (int i = 1; i <= n; ++i) sort(e[i].begin(), e[i].end());
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= deg[i]; ++j) {
      if (s[e[i][j - 1].x].find(state(deg[i], j)) != s[e[i][j - 1].x].end())
        ++vis[deg[i]][j];
      s[e[i][j - 1].x].insert(state(deg[i], j));
    }
  for (int i = 1; i <= n; ++i) {
    cn = 0;
    for (set<state>::iterator it = s[i].begin(); it != s[i].end(); ++it)
      now[++cn] = *it;
    for (int i = 1; i <= cn; ++i)
      for (int j = 1; j <= cn; ++j)
        b[now[i].x][now[i].y][now[j].x][now[j].y] = 1;
  }
  Dfs_ans(1);
  printf("%d\n", ans);
}
int main() {
  int T = 1;
  for (; T--;) work();
  return 0;
}
