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>
#pragma GCC optimize("O3")
using namespace std;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
const int inf = 1000000007;
const int MOD = 1000000007;
int num_memo[10][10];
vector<long long> ran;
long long hhash(int x) { return ran[x]; }
struct Node {
  int no;
  vector<pair<int, int>> edge;
  void build(int i) { no = i; }
};
int n, k;
long long ok = 0;
int calc(vector<int> const& status, vector<vector<long long>> const& memo) {
  long long sm = 0;
  long long cnt = 0;
  for (long long i = 0; i < (10); i++) {
    sm = sm ^ memo[i][status[i]];
    cnt += num_memo[i][status[i]];
  }
  if (sm == ok && cnt == n)
    return 1;
  else
    return 0;
}
int rec(vector<int> status, vector<vector<long long>> const& memo, int cur) {
  if (cur == k + 1) return calc(status, memo);
  int res = 0;
  for (long long i = 1; i <= cur; i++) {
    status[cur] = i;
    res += rec(status, memo, cur + 1);
  }
  return res;
}
void output(vector<vector<long long>> const& memo) {
  for (long long i = 1; i <= k; i++) {
    for (long long j = 1; j <= k; i++) {
      cout << memo[i][j] << " ";
    }
    cout << '\n';
  }
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  ran.resize(200005);
  for (long long i = 0; i < (200005); i++) {
    mt19937 mt(i);
    ran[i] = mt() % MOD;
  }
  for (long long i = 0; i < (10); i++)
    for (long long j = 0; j < (10); j++) num_memo[i][j] = 0;
  int m;
  cin >> n >> m >> k;
  vector<Node> node(n + 1);
  for (long long i = 1; i <= n; i++) node[i].build(i);
  for (long long i = 0; i < (m); i++) {
    int a, b, l;
    cin >> a >> b >> l;
    node[a].edge.push_back({l, b});
  }
  vector<vector<vector<int>>> in(n + 1,
                                 vector<vector<int>>(10, vector<int>(10, 0)));
  vector<vector<long long>> memo(10, vector<long long>(10, 0));
  for (long long i = 1; i <= n; i++) {
    sort(node[i].edge.begin(), node[i].edge.end());
    int num = node[i].edge.size();
    for (long long j = 0; j < (num); j++) {
      memo[num][j + 1] = memo[num][j + 1] ^ hhash(node[i].edge[j].second);
      in[node[i].edge[j].second][num][j + 1] += 1;
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long x = 0; x < (10); x++) {
      for (long long y = 0; y < (10); y++) {
        num_memo[x][y] = num_memo[x][y] + in[i][x][y];
      }
    }
  }
  for (long long i = 1; i <= n; i++) {
    ok = ok ^ (hhash(i));
  }
  vector<int> status(10, 1);
  cout << rec(status, memo, 1) << '\n';
  return 0;
}
