<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

476. Coach's Trouble
Time limit per test: 0.5 second(s)
Memory limit: 65536 kilobytes
input: standard
output: standard

Berland is a modern country, and one could hardly find a man, who would believe evil spirits. Everybody knows that celebrating Halloween is just tribute to old tradition. However, a coach of Berland University student programming teams wants to divide students (3N in total) into teams of three to have as many demonic teams as possible at the moment of future Halloween contest. But he knows for some triples of students that they can't make a demonic team, as they haven't performed well in action in previous contests. He has a list of these K forbidden triples. The coach supposes that any three students can make a demonic team unless they do not form a triple contained in the list. And now he wants to know the number of such partitions that all teams are demonic.

Input
The first line of the input contains two integers N and K, separated by one space (1 ≤ N ≤ 1000, 0 ≤ K ≤ 20). Next K lines contain three integers each ai, bi, ci (1 ≤ i ≤ K, 1 ≤ ai, bi, ci ≤ 3N). All triples are unique, that is they all are diffent as sets, and ai ≠q bi, ai ≠q ci, bi ≠q ci.

Output
The output should contain the only number without leading zeroes — the answer to the task.

Example(s)
sample input
sample output
2 0
10

sample input
sample output
2 3
1 2 3
4 5 6
1 4 5
8

<|response|>
1. Abridged Problem Statement  
We have 3·N labeled students (numbered 1…3N). We want to form N unordered teams of size 3 (both the teams and the order inside each team are unlabeled), so that none of the formed teams appears in a given list of K forbidden triples. Compute the exact number of valid partitions.  
Constraints:  
• 1 ≤ N ≤ 1000  
• 0 ≤ K ≤ 20  
• Forbidden triples are all distinct and consist of three different student indices.

2. Key Observations  
- The total number of ways to partition 3N labeled items into N unlabeled triplets is  
  total(N) = (3N)! / ( (3!)^N · N! ).  
- We need to exclude any partition that uses at least one forbidden triple. Since K ≤ 20, we can apply the inclusion–exclusion principle over the set of forbidden triples.  
- For any subset S of t forbidden triples that are pairwise disjoint, the number of partitions that include *all* triples in S is  
  ways(S) = (3(N–t))! / ( (3!)^(N–t) · (N–t)! ).  
- By inclusion–exclusion, the final answer is  
  Σ_{S⊆F} (–1)^{|S|} · ways(S),  
  where we drop any S whose triples overlap (they contribute zero).  
- We can implement the sum over all 2^K subsets by backtracking on the forbidden triples, keeping a “used” array of size 3N to detect overlaps, and accumulating a coefficient coef[c] for c = 0,3,6,…,3K equal to the sum of (–1)^{|S|} over all S covering c students.

3. Full Solution Approach  
a. Precompute factorials fact[i] = i! for i = 0…3N, and power6[j] = 6^j for j = 0…N.  
b. Maintain an array used[1…3N] initially all false. Maintain coef[0…3N] initially all zero.  
c. Define a recursive function backtrack(pos, covered, sign):  
   - pos is the index of the next forbidden triple to consider (0…K).  
   - covered is the total number of students covered so far = 3·(number of chosen triples).  
   - sign is +1 or –1 = (–1)^(#chosen so far).  
   • If pos == K, do coef[covered] += sign and return.  
   • First, skip forbidden[pos]: call backtrack(pos+1, covered, sign).  
   • Then, try to include forbidden[pos]: check its three students; if none are marked used[], mark them, recurse with backtrack(pos+1, covered+3, –sign), then unmark them.  
d. After backtracking, we have coef[c] for each c multiple of 3. For each c with coef[c] ≠ 0, let t = c/3. The number of ways to partition the remaining 3(N–t) students is  
   fact[3(N–t)] / ( power6[N–t] · (N–t)! ).  
   Multiply by coef[c] and add to the answer.  
e. Output the final answer as a big integer.

4. C++ Implementation with Detailed Comments  
```cpp
#include <bits/stdc++.h>
#include <boost/multiprecision/cpp_int.hpp>
using namespace std;
using BigInt = boost::multiprecision::cpp_int;

// Globals for simplicity
int N, K;
vector<array<int,3>> forbidden;   // forbidden triples (0-based indices)
vector<bool> used;                // used[i] = true if student i is already covered
vector<int> coef;                 // coef[c] accumulates inclusion–exclusion sums
vector<BigInt> fact, power6;      // factorials and 6-powers

// Backtracking over forbidden triples
// pos     = current index in [0..K)
// covered = number of students covered so far = 3*(#triples chosen)
// sign    = +1 or -1 = (–1)^(#triples chosen so far)
void backtrack(int pos, int covered, int sign) {
    if (pos == K) {
        // Record the coefficient for this covered count
        coef[covered] += sign;
        return;
    }
    // 1) Skip this forbidden triple
    backtrack(pos + 1, covered, sign);

    // 2) Try to include this forbidden triple, if no student overlap
    auto &tri = forbidden[pos];
    bool conflict = false;
    vector<int> marked;
    for (int x : tri) {
        if (used[x]) {
            conflict = true;
            break;
        }
        used[x] = true;
        marked.push_back(x);
    }
    if (!conflict) {
        // Flip sign when we include one more triple
        backtrack(pos + 1, covered + 3, -sign);
    }
    // Undo the 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++) {
        int a, b, c;
        cin >> a >> b >> c;
        // convert to 0-based
        forbidden[i] = {a-1, b-1, c-1};
    }

    // Initialize used array and coef array
    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 inclusion–exclusion coefficients
    backtrack(0, 0, +1);

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

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

5. Python Implementation with Detailed Comments  
```python
import sys
sys.set_int_max_str_digits(10**7)  # For very large ints in Python 3.11+

def main():
    input_data = sys.stdin.read().split()
    n, k = map(int, input_data[:2])
    triples = []
    idx = 2
    for _ in range(k):
        a, b, c = map(int, input_data[idx:idx+3])
        triples.append((a-1, b-1, c-1))  # convert to 0-based
        idx += 3

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

    used = [False] * (3*n)
    coef = [0] * (3*n + 1)

    # Recursively build inclusion–exclusion coefficients
    def backtrack(pos, covered, sign):
        if pos == k:
            coef[covered] += sign
            return
        # 1) skip this triple
        backtrack(pos+1, covered, sign)
        # 2) include this triple if no overlap
        tri = triples[pos]
        conflict = False
        marked = []
        for x in tri:
            if used[x]:
                conflict = True
                break
            used[x] = True
            marked.append(x)
        if not conflict:
            backtrack(pos+1, covered+3, -sign)
        # undo
        for x in marked:
            used[x] = False

    backtrack(0, 0, 1)

    # Compute final answer
    ans = 0
    for covered, c in enumerate(coef):
        if c == 0:
            continue
        # covered = 3*t
        t = covered // 3
        ways = fact[3*(n-t)] // (power6[n-t] * fact[n-t])
        ans += c * ways

    print(ans)

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