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 long long N = 2e5 + 5;
const long long K = 12;
vector<pair<long long, long long> > g[N];
long long mod[3] = {998244353, (long long)(1e9) + 7, (long long)(1e9) + 9}, res;
struct hash_num {
  long long a[3];
  hash_num() {
    a[0] = 0;
    a[1] = 1;
    a[2] = 0;
  }
  hash_num(long long x) {
    for (long long i = 0; i < 3; i++) a[i] = x % mod[i];
  }
  hash_num operator+(hash_num x) {
    hash_num res;
    res.a[0] = (a[0] + x.a[0]) % mod[0];
    res.a[1] = (a[1] * x.a[1]) % mod[1];
    res.a[2] = (a[2] + x.a[2]) % mod[2];
    return res;
  }
  bool operator==(hash_num x) {
    for (long long i = 0; i < 3; i++)
      if (a[i] != x.a[i]) return 0;
    return 1;
  }
};
hash_num sum, c[K][K], val[N];
long long rng() {
  long long a = rand();
  for (long long i = 0; i < 2; i++) a <<= 15, a ^= rand();
  return a;
}
void dfs(long long cur, long long k, hash_num x) {
  if (cur == k + 1) {
    if (x == sum) res++;
    return;
  }
  for (long long i = 1; i <= cur; i++) dfs(cur + 1, k, x + c[cur][i]);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  srand(time(NULL));
  long long n, m, k;
  cin >> n >> m >> k;
  while (m--) {
    long long u, v, w;
    cin >> u >> v >> w;
    g[u].push_back({w, v});
  }
  for (long long i = 1; i <= n; i++)
    val[i] = hash_num(rng()), sum = sum + val[i];
  for (long long u = 1; u <= n; u++) {
    sort(g[u].begin(), g[u].end());
    for (long long i = 1; i <= g[u].size(); i++)
      c[g[u].size()][i] = c[g[u].size()][i] + val[g[u][i - 1].second];
  }
  hash_num st;
  dfs(1, k, st);
  cout << res;
  return 0;
}
