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 K = 10;
int ans, locked[K][K];
vector<pair<int, int> > locks[K][K];
void search(int i) {
  if (i < 1) {
    ans++;
  } else {
    for (int j = 1; j <= i; j++) {
      if (locked[i][j] > 0) continue;
      for (auto [x, y] : locks[i][j]) {
        locked[x][y]++;
      }
      search(i - 1);
      for (auto [x, y] : locks[i][j]) {
        locked[x][y]--;
      }
    }
  }
}
struct Edge {
  int from, to, pos;
};
int main() {
  int n, m, k;
  scanf("%d %d %d", &n, &m, &k);
  vector<Edge> e(m + 1);
  for (int j = 1; j <= m; j++) {
    Edge a;
    int i;
    scanf("%d %d %d", &a.from, &a.to, &i);
    e[i] = a;
  }
  vector<int> out(n + 1);
  vector<vector<int> > in(n + 1);
  for (int i = 1; i <= m; i++) {
    out[e[i].from]++;
    e[i].pos = out[e[i].from];
    in[e[i].to].push_back(i);
  }
  for (int v = 1; v <= n; v++) {
    set<pair<int, int> > single;
    for (int i : in[v]) {
      pair<int, int> ass(out[e[i].from], e[i].pos);
      if (single.count(ass)) {
        locked[ass.first][ass.second]++;
      } else {
        single.insert(ass);
      }
    }
    for (auto it = single.begin(); it != single.end(); it++) {
      for (auto jt = single.begin(); jt != single.end(); jt++) {
        if (it == jt) continue;
        locks[it->first][it->second].push_back(*jt);
      }
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 1; j <= i; j++) {
      auto& x = locks[i][j];
      sort(x.begin(), x.end());
      x.erase(unique(x.begin(), x.end()), x.end());
    }
  }
  search(k);
  printf("%d\n", ans);
}
