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 maxn = 2e5 + 9;
const long long T = 12;
bool kek[T][T][T][T];
vector<pair<long long, long long> > g[maxn];
vector<pair<long long, long long> > q[maxn];
long long n, m, k;
long long ans;
void gen(vector<long long>& cur) {
  if ((long long)cur.size() == k) {
    ans++;
    return;
  }
  long long x = (long long)cur.size() + 1;
  for (long long i = 0; i < x; i++) {
    if (kek[x][i][x][i]) continue;
    bool fl = 1;
    for (long long e = 0; e < (long long)cur.size(); e++) {
      if (kek[e + 1][cur[e]][x][i]) {
        fl = 0;
        break;
      }
    }
    if (fl) {
      cur.emplace_back(i);
      gen(cur);
      cur.pop_back();
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  for (long long i = 0; i < m; i++) {
    long long a, b, c;
    cin >> a >> b >> c;
    a--, b--;
    g[a].emplace_back(c, b);
  }
  for (long long v = 0; v < n; v++) {
    sort(g[v].begin(), g[v].end());
    long long x = g[v].size();
    for (long long i = 0; i < x; i++) q[g[v][i].second].emplace_back(x, i);
  }
  for (long long v = 0; v < n; v++) {
    sort(q[v].begin(), q[v].end());
    for (long long i = 1; i < (long long)q[v].size(); i++) {
      if (q[v][i] == q[v][i - 1])
        kek[q[v][i].first][q[v][i].second][q[v][i].first][q[v][i].second] = 1;
    }
    sort(q[v].begin(), q[v].end()),
        q[v].resize(unique(q[v].begin(), q[v].end()) - q[v].begin());
    for (long long i = 0; i < (long long)q[v].size(); i++)
      for (long long e = 0; e < (long long)q[v].size(); e++) {
        if (e == i) continue;
        kek[q[v][i].first][q[v][i].second][q[v][e].first][q[v][e].second] = 1;
      }
  }
  vector<long long> cur;
  gen(cur);
  cout << ans;
}
