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;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using ld = long double;
const int MN = 2e5 + 5, MOD = 1e9 + 7;
vector<pii> adj[MN];
bitset<MN> bb[10][10];
unordered_map<bitset<MN>, int> cntA, cntB;
int deg[MN];
int main() {
  int n, m, k;
  scanf("%d %d %d", &n, &m, &k);
  while (m--) {
    int a, b, c;
    scanf("%d %d %d", &a, &b, &c);
    adj[a].emplace_back(b, c);
    ++deg[a];
  }
  for (int i = 1; i <= n; i++) {
    sort((adj[i]).begin(), (adj[i]).end(),
         [](pii a, pii b) { return a.second < b.second; });
    for (int j = 0; j < deg[i]; j++) bb[deg[i]][j + 1].set(adj[i][j].first);
  }
  bitset<MN> mask;
  for (int i = 1; i <= n; i++) mask.set(i);
  function<void(int, bitset<MN> &)> gen = [&](int cur, bitset<MN> &bs) {
    if (cur > k || cur > 6) {
      cntA[bs]++;
      return;
    }
    for (int take = 1; take <= cur; take++)
      if (!(bs & bb[cur][take]).count()) {
        bs |= bb[cur][take];
        gen(cur + 1, bs);
        bs &= (mask ^ bb[cur][take]);
      }
  };
  function<void(int, bitset<MN> &)> gen2 = [&](int cur, bitset<MN> &bs) {
    if (cur > k) {
      cntB[bs]++;
      return;
    }
    for (int take = 1; take <= cur; take++)
      if (!(bs & bb[cur][take]).count()) {
        bs |= bb[cur][take];
        gen2(cur + 1, bs);
        bs &= (mask ^ bb[cur][take]);
      }
  };
  bitset<MN> bs;
  gen(1, bs);
  if (k <= 6) {
    int ret = 0;
    for (auto &au : cntA)
      if (au.first == mask) ret += au.second;
    return !printf("%d\n", ret);
  }
  assert(!bs.count());
  gen2(7, bs);
  ll ret = 0;
  for (auto &au : cntB) {
    bitset<MN> need = mask ^ au.first;
    if (cntA.count(need)) ret += (ll)au.second * cntA[need];
  }
  printf("%lld\n", ret);
  return 0;
}
