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 inf = 0x3f3f3f3f;
const long long maxn = 201110;
const long long M = 1e9 + 7;
long long n, m, k, ans;
unsigned long long p[maxn], all, a[15][15];
unsigned long long mo = 1e9 + 7;
vector<pair<long long, long long> > v[maxn];
void dfs(long long u, unsigned long long hash) {
  if (u > k) {
    ans += (hash == all);
    return;
  }
  for (long long i = 1; i <= u; i++) {
    dfs(u + 1, hash + a[u][i]);
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> k;
  for (long long i = 1; i <= m; i++) {
    long long x, y, z;
    cin >> x >> y >> z;
    v[x].push_back({z, y});
  }
  p[0] = 1;
  for (long long i = 1; i <= n; i++) {
    p[i] = p[i - 1] * mo;
    all += p[i];
  }
  for (long long i = 1; i <= n; i++) {
    sort(v[i].begin(), v[i].end());
    long long sz = v[i].size();
    for (long long j = 0; j < sz; j++) {
      a[sz][j + 1] += p[v[i][j].second];
    }
  }
  dfs(1, 0);
  cout << ans << endl;
  return 0;
}
