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>
template <typename T>
inline void read(T &x) {
  char c;
  bool nega = 0;
  while ((!isdigit(c = getchar())) && (c != '-'))
    ;
  if (c == '-') {
    nega = 1;
    c = getchar();
  }
  x = c - 48;
  while (isdigit(c = getchar())) x = x * 10 + c - 48;
  if (nega) x = -x;
}
template <typename T>
inline void writep(T x) {
  if (x > 9) writep(x / 10);
  putchar(x % 10 + 48);
}
template <typename T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  writep(x);
  putchar(' ');
}
template <typename T>
inline void writeln(T x) {
  write(x);
  putchar('\n');
}
using namespace std;
const int base = 200005;
const int mod = 1000000007;
int po[200005], ok, h[15][15], ans;
int n, m, k;
vector<pair<int, int> > a[200005];
bool comp(vector<pair<int, int> > a, vector<pair<int, int> > b) {
  return a.size() < b.size();
}
void Dfs(int u, int check) {
  if (u == k) {
    ans += check == ok;
    return;
  }
  ++u;
  for (int i = (1); i <= (u); ++i) {
    Dfs(u, (check + h[u][i]) % mod);
  }
}
int main() {
  srand(time(NULL));
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie();
  if (fopen(".inp", "r")) freopen(".inp", "r", stdin);
  cin >> n >> m >> k;
  for (int i = (1); i <= (m); ++i) {
    int u, v, w;
    cin >> u >> v >> w;
    a[u].push_back(make_pair(w, v));
  }
  po[0] = 1;
  for (int i = (1); i <= (n); ++i) {
    po[i] = 1ll * po[i - 1] * base % mod;
    ok = (ok + po[i]) % mod;
  }
  for (int i = (1); i <= (n); ++i) {
    sort(a[i].begin(), a[i].end());
  }
  sort(a + 1, a + n + 1, comp);
  for (int i = (1); i <= (n); ++i) {
    int r = i;
    while (r <= n && a[r].size() == a[i].size()) ++r;
    --r;
    int temp = a[i].size();
    for (int j = (1); j <= (temp); ++j) {
      for (int hh = (i); hh <= (r); ++hh) {
        h[temp][j] = (h[temp][j] + po[a[hh][j - 1].second]) % mod;
      }
    }
    i = r;
  }
  Dfs(0, 0);
  cout << ans;
  return 0;
}
