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 zero = 0;
const long long INF = 3000000000000000000;
const int inINF = 1000000000;
const long long MOD = 1000000007;
const long long MOD2 = 998244353;
void Yes() { printf("Yes\n"); }
void No() { printf("No\n"); }
void YES() { printf("YES\n"); }
void NO() { printf("NO\n"); }
int n, m, k;
void next(vector<int> &c) {
  c[k - 1]++;
  bool check = true;
  int cnt = 1;
  while (check) {
    if (c[k - cnt] > k - cnt + 1) {
      if (k - cnt - 1 >= 0) {
        c[k - cnt - 1]++;
        c[k - cnt] = 1;
      } else {
        check = false;
      }
      cnt++;
    } else {
      check = false;
    }
  }
}
int main() {
  cin >> n >> m >> k;
  vector<vector<pair<int, int>>> G(n);
  int u, v, w;
  for (int i = 0; i < m; i++) {
    cin >> u >> v >> w;
    u--;
    v--;
    G[u].push_back(make_pair(w, v));
  }
  int dim = k * k * k * k;
  vector<int> a(dim, true);
  vector<set<pair<int, int>>> H(n);
  int s, cnt;
  for (int i = 0; i < n; i++) {
    sort((G[i]).begin(), (G[i]).end());
    s = G[i].size();
    cnt = 1;
    for (pair<int, int> x : G[i]) {
      if (H[x.second].count(make_pair(s, cnt))) {
        a[k * k * k * (s - 1) + k * k * (cnt - 1) + k * (s - 1) + (cnt - 1)] =
            false;
      }
      H[x.second].insert(make_pair(s, cnt));
      cnt++;
    }
  }
  int cnt2;
  for (int i = 0; i < n; i++) {
    cnt = 0;
    for (pair<int, int> x : H[i]) {
      cnt2 = 0;
      for (pair<int, int> y : H[i]) {
        if (cnt == cnt2) {
          cnt2++;
          continue;
        }
        a[k * k * k * (x.first - 1) + k * k * (x.second - 1) +
          k * (y.first - 1) + (y.second - 1)] = false;
        cnt2++;
      }
      cnt++;
    }
  }
  vector<int> c(k, 1);
  vector<long long> fact(10);
  fact[0] = 1;
  fact[1] = 1;
  for (int i = 2; i < 10; i++) {
    fact[i] = fact[i - 1] * i % MOD;
  }
  long long ans = fact[k];
  bool check;
  for (int i = 0; i < fact[k]; i++) {
    check = true;
    for (int j = 0; j < k; j++) {
      for (int l = 0; l < k; l++) {
        if (a[k * k * k * j + k * k * (c[j] - 1) + k * l + (c[l] - 1)] ==
            false) {
          ans--;
          check = false;
          break;
        }
      }
      if (check == false) {
        break;
      }
    }
    next(c);
  }
  printf("%lld\n", ans);
}
