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;
using ll = long long;
using P = pair<int, int>;
vector<int> cis(9);
int ans = 0;
void dfs(int now, int k, vector<vector<ll>> &hashs, ll ch) {
  int i;
  if (now == k) {
    ll comp = 0;
    for (int i = 0; i < k; i++) {
      comp = (comp + hashs[i][cis[i]]) % 998244353;
    }
    if (comp == ch) ans++;
    return;
  } else {
    for (int i = 0; i < now + 1; i++) {
      cis[now] = i;
      dfs(now + 1, k, hashs, ch);
    }
  }
}
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> u(m);
  vector<int> v(m);
  vector<int> w(m);
  int i, j;
  for (int i = 0; i < m; i++) cin >> u[i] >> v[i] >> w[i];
  for (int i = 0; i < m; i++) {
    u[i]--;
    v[i]--;
  }
  vector<vector<P>> e(n);
  for (int i = 0; i < m; i++) {
    e[u[i]].emplace_back(make_pair(w[i], v[i]));
  }
  vector<vector<ll>> hashs(k, vector<ll>(k));
  vector<ll> vhash(n, 1);
  int ks = 1000000;
  for (int i = 0; i < n - 1; i++) {
    vhash[i + 1] = vhash[i] * ks % 998244353;
  }
  for (int i = 0; i < n; i++) {
    int sz = e[i].size();
    priority_queue<P, vector<P>, greater<P>> pq;
    for (int j = 0; j < sz; j++) {
      pq.push(e[i][j]);
    }
    for (int j = 0; j < sz; j++) {
      P tops = pq.top();
      pq.pop();
      int vnum = tops.second;
      hashs[sz - 1][j] = (hashs[sz - 1][j] + vhash[vnum]) % 998244353;
    }
  }
  ll ch = 0;
  for (int i = 0; i < n; i++) {
    ch = (ch + vhash[i]) % 998244353;
  }
  dfs(0, k, hashs, ch);
  cout << ans << endl;
}
