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 + 7;
long long mod[3];
struct hash_n {
  long long a[3];
  hash_n() { a[0] = a[1] = a[2] = 0; }
  hash_n(long long x) {
    for (int i = 0; i < 3; i++) {
      a[i] = (x % mod[i] + mod[i]) % mod[i];
    }
  }
  bool operator==(const hash_n &x) const {
    return a[0] == x.a[0] && a[1] == x.a[1] && a[2] == x.a[2];
  }
  hash_n operator+(hash_n x) {
    x.a[0] += a[0];
    x.a[0] %= mod[0];
    x.a[1] *= a[1];
    x.a[1] %= mod[1];
    x.a[2] += a[2];
    x.a[2] %= mod[2];
    return x;
  }
} h[N], s, d[10][10];
vector<pair<int, int>> g[N];
int res = 0;
void calc(int i, int k, hash_n x) {
  if (i == k) {
    if (x == s) {
      res++;
    }
    return;
  }
  for (int j = 0; j <= i; j++) {
    calc(i + 1, k, x + d[i + 1][j + 1]);
  }
}
void solve() {
  mod[0] = 998244353;
  mod[1] = 1e9 + 7;
  mod[2] = (rand()) % (mod[0]) + 1e8;
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v, w;
    cin >> u >> v >> w;
    g[u].push_back(make_pair(w, v));
  }
  for (int i = 1; i <= n; i++) {
    h[i] = hash_n(rand() / 2 * rand());
    s = s + h[i];
    sort(g[i].begin(), g[i].end());
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < g[i].size(); j++) {
      d[g[i].size()][j + 1] = d[g[i].size()][j + 1] + h[g[i][j].second];
    }
  }
  calc(0, k, hash_n());
  cout << res << endl;
}
void t(int t) {
  if (!t) {
    cin >> t;
  }
  while (t--) {
    solve();
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  t(1);
  return 0;
}
