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 = 2e5 + 3;
struct E {
  int v, w;
};
vector<E> out[N];
int n, m, k;
int bad[100][100], invalid[100];
set<int> conflit[N];
int chash(int d, int k) { return d * 10 + k; }
pair<int, int> ihash(int x) { return {x / 10, x % 10}; }
int dfs() {
  static vector<int> a;
  int na = a.size(), d = na + 1, cnt = 0;
  if (na == k) return 1;
  for (int i = 1; i <= d; i++) {
    int failed = 0;
    auto x = chash(d, i);
    if (invalid[x]) continue;
    for (int j = 0; j < na; j++) {
      if (bad[x][chash(j + 1, a[j])]) {
        failed = 1;
        break;
      }
    }
    if (!failed) {
      a.push_back(i);
      cnt += dfs();
      a.pop_back();
    }
  }
  return cnt;
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  for (int i = 0, u, v, w; i < m; i++) {
    scanf("%d%d%d", &u, &v, &w);
    out[u].push_back({v, w});
  }
  for (int u = 1; u <= n; u++) {
    sort(out[u].begin(), out[u].end(), [](E& l, E& r) { return l.w < r.w; });
    for (int i = 0; i < out[u].size(); i++) {
      int v = out[u][i].v, d = out[u].size();
      int h = chash(d, i + 1);
      if (conflit[v].find(h) != conflit[v].end())
        invalid[h] = 1;
      else
        conflit[v].insert(h);
    }
  }
  for (int u = 1; u <= n; u++)
    if (conflit[u].size() > 1)
      for (auto x : conflit[u])
        for (auto y : conflit[u]) bad[x][y] = 1;
  printf("%d", dfs());
  return 0;
}
