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 MAXN = 2e5 + 7;
vector<vector<pair<long long, long long>>> g(MAXN);
long long k;
vector<long long> p;
long long d[MAXN];
long long deg[MAXN];
long long H[10][10];
long long r[MAXN];
long long ans = 0;
long long sum = 0;
void gen(long long i) {
  if (i == k) {
    long long cur = 0;
    for (long long i = 0; i < k; ++i) {
      cur += H[i + 1][p[i] + 1];
    }
    if (cur == sum) {
      ++ans;
    }
    return;
  }
  for (long long j = 0; j <= i; ++j) {
    p.push_back(j);
    gen(i + 1);
    p.pop_back();
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  srand(clock());
  long long n, m;
  cin >> n >> m >> k;
  for (long long i = 0; i < m; ++i) {
    long long u, v, w;
    cin >> u >> v >> w;
    --u;
    --v;
    g[u].push_back({w, v});
    ++d[u];
  }
  for (long long i = 0; i < n; ++i) {
    sort(g[i].begin(), g[i].end());
  }
  for (long long i = 0; i < n; ++i) {
    r[i] = rand() * 32 + rand();
    sum += r[i];
  }
  for (long long i = 1; i <= k; ++i) {
    for (long long j = 1; j <= i; ++j) {
      for (long long u = 0; u < n; ++u) {
        if (d[u] == i) {
          H[i][j] += r[g[u][j - 1].second];
        }
      }
    }
  }
  gen(0);
  cout << ans << endl;
  return 0;
}
