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;
vector<pair<long long, long long> > adj[200010];
long long hash1[10][10], hash2[10][10], val[200010];
vector<long long> pres;
long long kk;
long long cnt = 0, ch1 = 0, ch2 = 1;
long long mod = 1000000067;
void solve(long long ind) {
  if (ind > kk) {
    long long sum1 = 0, sum2 = 1;
    for (long long i = 0; i < kk; i++) {
      long long y = pres[i];
      sum1 += hash1[i + 1][y];
      sum2 *= hash2[i + 1][y];
      sum2 %= mod;
    }
    if (sum1 == ch1 && sum2 == ch2) cnt++;
    return;
  }
  for (long long i = 1; i <= ind; i++) {
    pres.push_back(i);
    solve(ind + 1);
    pres.pop_back();
  }
}
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m, k;
  cin >> n >> m >> k;
  kk = k;
  for (long long i = 0; i < m; i++) {
    long long a, b, w;
    cin >> a >> b >> w;
    adj[a].push_back(make_pair(w, b));
  }
  for (long long i = 0; i < 10; i++) {
    for (long long j = 0; j < 10; j++) {
      hash2[i][j] = 1;
    }
  }
  for (long long i = 1; i <= n; i++) {
    ch1 += i * i;
    ch2 *= i;
    ch2 %= mod;
    sort(adj[i].begin(), adj[i].end());
  }
  for (long long i = 1; i <= n; i++) {
    long long s = adj[i].size();
    for (long long j = 0; j < s; j++) {
      long long x = adj[i][j].second;
      hash1[s][j + 1] += x * x;
      hash2[s][j + 1] *= x;
      hash2[s][j + 1] %= mod;
    }
  }
  solve(1);
  cout << cnt;
}
