1. Abridged Problem Statement  
Given 3N labeled students (numbered 1…3N), we want to count the number of ways to split them into N unordered teams of 3 (teams themselves are unlabeled, and within each team the order does not matter), subject to the constraint that none of the chosen teams is one of K forbidden triples.  N≤1000, K≤20.  Output the exact count.

2. Detailed Editorial  

Let total(N) be the number of ways to partition 3N items into N unlabeled triples.  It is  
  total(N) = (3N)!  / ( (3!)^N · N! ).  

We have a set F of K forbidden triples.  We wish to count partitions that include *no* forbidden triple.  By the principle of inclusion–exclusion, if we let for any subset S⊆F the quantity  
  ways(S) = number of partitions that *do* include every triple of S  
(only nonzero if the triples in S are disjoint), then the answer is  
  Σ_{S⊆F}  (–1)^{|S|} · ways(S).  

If |S|=t and its t triples are disjoint, those triples cover 3t distinct students, so the remaining 3(N–t) students can be arbitrarily partitioned into N–t triples.  Hence  
  ways(S) = (3(N–t))!  / ( (3!)^{N–t} · (N–t)! ).  

Summing over all 2^K subsets S (and dropping those that overlap), we get a direct inclusion–exclusion.  
Because K≤20, 2^K is about 1 048 576, which is feasible.  Precompute factorials up to 3N, and powers of 6 up to N, so each term can be evaluated in O(1) big-integer operations.  To generate the valid subsets S and keep track of overlaps efficiently, we do a backtracking over the K forbidden triples, maintaining a “used” marker for each student to detect conflicts, and accumulate a coefficient coef[cnt] = Σ_{S:|S|·3=cnt} (–1)^{|S|}.  Finally, answer = Σ_{cnt} coef[cnt] · (3N–cnt)! / (6^{N–cnt/3} · (N–cnt/3)!).

3. Provided C++ Solution (with detailed comments)

```cpp
#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>
using namespace std;
using BigInt = boost::multiprecision::cpp_int;

// Global containers (to avoid passing by reference everywhere)
int N, K;
vector<array<int,3>> forbidden;   // list of forbidden triples
vector<bool> used;                // marks which students are used in current subset
vector<int> coef;                 // coef[cnt] accumulates inclusion–exclusion sign sums
vector<BigInt> fact;              // factorials up to 3N
vector<BigInt> power6;            // powers of 6 up to N

// Recursively choose a subset of forbidden triples
// pos = current index in forbidden[]
// cnt = total number of students covered so far = 3 * (# of chosen triples)
// sign = (+1) or (–1) depending on parity of chosen triples
void backtrack(int pos, int cnt, int sign) {
    if (pos == K) {
        // we've decided for every triple whether to include it
        coef[cnt] += sign;
        return;
    }
    // Option 1: skip forbidden[pos]
    backtrack(pos + 1, cnt, sign);

    // Option 2: include forbidden[pos], if it doesn't overlap
    bool conflict = false;
    array<int,3> &tri = forbidden[pos];
    vector<int> marked;  // track which students we mark, to undo later
    for (int x: tri) {
        if (used[x]) { conflict = true; break; }
        used[x] = true;
        marked.push_back(x);
    }
    if (!conflict) {
        // no overlap: we can safely include this triple
        backtrack(pos + 1, cnt + 3, -sign);
    }
    // undo marks
    for (int x: marked) {
        used[x] = false;
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    // Read input
    cin >> N >> K;
    forbidden.resize(K);
    for (int i = 0; i < K; i++) {
        cin >> forbidden[i][0]
            >> forbidden[i][1]
            >> forbidden[i][2];
        // convert to 0-based for easy indexing
        forbidden[i][0]--;
        forbidden[i][1]--;
        forbidden[i][2]--;
    }

    // Prepare space
    used.assign(3*N, false);
    coef.assign(3*N + 1, 0);

    // Precompute factorials up to 3N
    fact.assign(3*N + 1, 1);
    for (int i = 1; i <= 3*N; i++) {
        fact[i] = fact[i-1] * i;
    }
    // Precompute powers of 6 up to N
    power6.assign(N + 1, 1);
    for (int i = 1; i <= N; i++) {
        power6[i] = power6[i-1] * 6;
    }

    // Build coef[cnt] via inclusion–exclusion recursion
    backtrack(0, 0, +1);

    // Sum up the final answer
    BigInt answer = 0;
    for (int cnt = 0; cnt <= 3*N; cnt++) {
        if (coef[cnt] == 0) continue;
        // cnt must be multiple of 3, let t = cnt/3
        int t = cnt / 3;
        // ways to partition remaining 3(N–t) students:
        // (3(N–t))! / (6^(N–t) * (N–t)!)
        BigInt ways = fact[3*(N-t)];
        ways /= (power6[N-t] * fact[N-t]);
        // add with sign coef[cnt]
        answer += BigInt(coef[cnt]) * ways;
    }

    // Output
    cout << answer << "\n";
    return 0;
}
```

4. Python Solution (with detailed comments)

```python
import sys
sys.set_int_max_str_digits(100000)  # allow very large ints if Python 3.11+

def main():
    # Read input: N=number of teams, K=number of forbidden triples
    n, k = map(int, sys.stdin.readline().split())
    triplets = [tuple(map(int, sys.stdin.readline().split())) for _ in range(k)]

    # Precompute factorials[0..3n] and powers of 6 up to n
    factorial = [1] * (3*n + 1)
    power6 = [1] * (n + 1)
    for i in range(1, 3*n + 1):
        factorial[i] = factorial[i - 1] * i
    for i in range(1, n + 1):
        power6[i] = power6[i - 1] * 6

    # 'used' marks which student indices (0-based) are already included
    used = [False] * (3*n)
    # coef[cnt] will collect Σ_{S subset of forbidden} (–1)^|S|, where S covers cnt students
    coef = [0] * (3*n + 1)

    # Recursive inclusion–exclusion over forbidden triples
    def rec(pos, cnt, sign):
        # pos = index of next forbidden triple to consider
        # cnt = how many students are already covered by chosen triples
        # sign = +1 or –1
        if pos == k:
            coef[cnt] += sign
            return

        # 1) skip this triple
        rec(pos + 1, cnt, sign)

        # 2) try to include this triple, if no overlap
        tri = triplets[pos]
        conflict = False
        marked = []
        for x in tri:
            idx = x-1
            if used[idx]:
                conflict = True
                break
            used[idx] = True
            marked.append(idx)
        if not conflict:
            # flip sign when adding one triple
            rec(pos + 1, cnt + 3, -sign)
        # undo marks
        for idx in marked:
            used[idx] = False

    # build coef[]
    rec(0, 0, +1)

    # compute answer via sum over cnt=3*t
    ans = 0
    for cnt, c in enumerate(coef):
        if c == 0: 
            continue
        t = cnt // 3
        # ways to partition remaining = factorial[3(n–t)] / (6^(n–t) * (n–t)!)
        ways = factorial[3*(n-t)] // (power6[n-t] * factorial[n-t])
        ans += c * ways

    print(ans)

if __name__ == "__main__":
    main()
```

5. Compressed Editorial  
- Total partitions: (3N)!/((3!)^N·N!).  
- Exclude forbidden triples by inclusion–exclusion over K (≤20) sets.  
- For any subset S of t disjoint forbidden triples: count = (3(N–t))!/((3!)^(N–t)·(N–t)!).  
- Sum (–1)^t · that count, over all S.  
- Implement by backtracking with a “used” array to skip overlapping S.  
- Precompute factorials (up to 3N) and powers of 6 (up to N).  
- Final answer is ∑_{cnt=3t} coef[cnt]·(3N–cnt)!/(6^(N–t)·(N–t)!).