<|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 power_6[j] = 6^j for j = 0…N using the bigint struct (results far exceed 64 bits).
b. Maintain an array used[1…3N] initially all false. Maintain coef[0…3N] initially all zero.
c. Define a recursive function rec(pos, cnt, p):
   - pos is the index of the next forbidden triple to consider (0…K).
   - cnt is the total number of students covered so far = 3·(number of chosen triples).
   - p is +1 or –1 = (–1)^(#chosen so far).
   • If pos == K, do coef[cnt] += p and return.
   • First, skip forbidden[pos]: call rec(pos+1, cnt, p).
   • Then, try to include forbidden[pos]: check its three students; if none are marked used[], mark them, recurse with rec(pos+1, cnt+3, –p), 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)] / ( power_6[N–t] · (N–t)! ).
   Multiply by coef[c] and add to the answer.
e. Output the final answer as a big integer.

4. C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

// base and base_digits must be consistent
const int base = 1000000000;
const int base_digits = 9;

struct bigint {
    vector<int> z;
    int sign;

    bigint() : sign(1) {}

    bigint(long long v) { *this = v; }

    bigint(const string& s) { read(s); }

    void operator=(const bigint& v) {
        sign = v.sign;
        z = v.z;
    }

    void operator=(long long v) {
        sign = 1;
        if(v < 0) {
            sign = -1, v = -v;
        }
        z.clear();
        for(; v > 0; v = v / base) {
            z.push_back(v % base);
        }
    }

    bigint operator+(const bigint& v) const {
        if(sign == v.sign) {
            bigint res = v;

            for(int i = 0, carry = 0;
                i < (int)max(z.size(), v.z.size()) || carry; ++i) {
                if(i == (int)res.z.size()) {
                    res.z.push_back(0);
                }
                res.z[i] += carry + (i < (int)z.size() ? z[i] : 0);
                carry = res.z[i] >= base;
                if(carry) {
                    res.z[i] -= base;
                }
            }
            return res;
        }
        return *this - (-v);
    }

    bigint operator-(const bigint& v) const {
        if(sign == v.sign) {
            if(abs() >= v.abs()) {
                bigint res = *this;
                for(int i = 0, carry = 0; i < (int)v.z.size() || carry; ++i) {
                    res.z[i] -= carry + (i < (int)v.z.size() ? v.z[i] : 0);
                    carry = res.z[i] < 0;
                    if(carry) {
                        res.z[i] += base;
                    }
                }
                res.trim();
                return res;
            }
            return -(v - *this);
        }
        return *this + (-v);
    }

    void operator*=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = 0, carry = 0; i < (int)z.size() || carry; ++i) {
            if(i == (int)z.size()) {
                z.push_back(0);
            }
            long long cur = z[i] * (long long)v + carry;
            carry = (int)(cur / base);
            z[i] = (int)(cur % base);
            // asm("divl %%ecx" : "=a"(carry), "=d"(a[i]) : "A"(cur),
            // "c"(base));
        }
        trim();
    }

    bigint operator*(int v) const {
        bigint res = *this;
        res *= v;
        return res;
    }

    friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1) {
        int norm = base / (b1.z.back() + 1);
        bigint a = a1.abs() * norm;
        bigint b = b1.abs() * norm;
        bigint q, r;
        q.z.resize(a.z.size());

        for(int i = a.z.size() - 1; i >= 0; i--) {
            r *= base;
            r += a.z[i];
            int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
            int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
            int d = ((long long)s1 * base + s2) / b.z.back();
            r -= b * d;
            while(r < 0) {
                r += b, --d;
            }
            q.z[i] = d;
        }

        q.sign = a1.sign * b1.sign;
        r.sign = a1.sign;
        q.trim();
        r.trim();
        return make_pair(q, r / norm);
    }

    friend bigint sqrt(const bigint& a1) {
        bigint a = a1;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        int n = a.z.size();

        int firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int norm = base / (firstDigit + 1);
        a *= norm;
        a *= norm;
        while(a.z.empty() || a.z.size() % 2 == 1) {
            a.z.push_back(0);
        }

        bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
        firstDigit = (int)sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
        int q = firstDigit;
        bigint res;

        for(int j = n / 2 - 1; j >= 0; j--) {
            for(;; --q) {
                bigint r1 =
                    (r - (res * 2 * base + q) * q) * base * base +
                    (j > 0 ? (long long)a.z[2 * j - 1] * base + a.z[2 * j - 2]
                           : 0);
                if(r1 >= 0) {
                    r = r1;
                    break;
                }
            }
            res *= base;
            res += q;

            if(j > 0) {
                int d1 =
                    res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
                int d2 =
                    res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
                int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()] : 0;
                q = ((long long)d1 * base * base + (long long)d2 * base + d3) /
                    (firstDigit * 2);
            }
        }

        res.trim();
        return res / norm;
    }

    bigint operator/(const bigint& v) const { return divmod(*this, v).first; }

    bigint operator%(const bigint& v) const { return divmod(*this, v).second; }

    void operator/=(int v) {
        if(v < 0) {
            sign = -sign, v = -v;
        }
        for(int i = (int)z.size() - 1, rem = 0; i >= 0; --i) {
            long long cur = z[i] + rem * (long long)base;
            z[i] = (int)(cur / v);
            rem = (int)(cur % v);
        }
        trim();
    }

    bigint operator/(int v) const {
        bigint res = *this;
        res /= v;
        return res;
    }

    int operator%(int v) const {
        if(v < 0) {
            v = -v;
        }
        int m = 0;
        for(int i = z.size() - 1; i >= 0; --i) {
            m = (z[i] + m * (long long)base) % v;
        }
        return m * sign;
    }

    void operator+=(const bigint& v) { *this = *this + v; }
    void operator-=(const bigint& v) { *this = *this - v; }
    void operator*=(const bigint& v) { *this = *this * v; }
    void operator/=(const bigint& v) { *this = *this / v; }

    bool operator<(const bigint& v) const {
        if(sign != v.sign) {
            return sign < v.sign;
        }
        if(z.size() != v.z.size()) {
            return z.size() * sign < v.z.size() * v.sign;
        }
        for(int i = z.size() - 1; i >= 0; i--) {
            if(z[i] != v.z[i]) {
                return z[i] * sign < v.z[i] * sign;
            }
        }
        return false;
    }

    bool operator>(const bigint& v) const { return v < *this; }
    bool operator<=(const bigint& v) const { return !(v < *this); }
    bool operator>=(const bigint& v) const { return !(*this < v); }
    bool operator==(const bigint& v) const {
        return !(*this < v) && !(v < *this);
    }
    bool operator!=(const bigint& v) const { return *this < v || v < *this; }

    void trim() {
        while(!z.empty() && z.back() == 0) {
            z.pop_back();
        }
        if(z.empty()) {
            sign = 1;
        }
    }

    bool isZero() const { return z.empty() || (z.size() == 1 && !z[0]); }

    bigint operator-() const {
        bigint res = *this;
        res.sign = -sign;
        return res;
    }

    bigint abs() const {
        bigint res = *this;
        res.sign *= res.sign;
        return res;
    }

    long long longValue() const {
        long long res = 0;
        for(int i = z.size() - 1; i >= 0; i--) {
            res = res * base + z[i];
        }
        return res * sign;
    }

    friend bigint gcd(const bigint& a, const bigint& b) {
        return b.isZero() ? a : gcd(b, a % b);
    }
    friend bigint lcm(const bigint& a, const bigint& b) {
        return a / gcd(a, b) * b;
    }

    void read(const string& s) {
        sign = 1;
        z.clear();
        int pos = 0;
        while(pos < (int)s.size() && (s[pos] == '-' || s[pos] == '+')) {
            if(s[pos] == '-') {
                sign = -sign;
            }
            ++pos;
        }
        for(int i = s.size() - 1; i >= pos; i -= base_digits) {
            int x = 0;
            for(int j = max(pos, i - base_digits + 1); j <= i; j++) {
                x = x * 10 + s[j] - '0';
            }
            z.push_back(x);
        }
        trim();
    }

    friend istream& operator>>(istream& stream, bigint& v) {
        string s;
        stream >> s;
        v.read(s);
        return stream;
    }

    friend ostream& operator<<(ostream& stream, const bigint& v) {
        if(v.sign == -1) {
            stream << '-';
        }
        stream << (v.z.empty() ? 0 : v.z.back());
        for(int i = (int)v.z.size() - 2; i >= 0; --i) {
            stream << setw(base_digits) << setfill('0') << v.z[i];
        }
        return stream;
    }

    static vector<int> convert_base(
        const vector<int>& a, int old_digits, int new_digits
    ) {
        vector<long long> p(max(old_digits, new_digits) + 1);
        p[0] = 1;
        for(int i = 1; i < (int)p.size(); i++) {
            p[i] = p[i - 1] * 10;
        }
        vector<int> res;
        long long cur = 0;
        int cur_digits = 0;
        for(int i = 0; i < (int)a.size(); i++) {
            cur += a[i] * p[cur_digits];
            cur_digits += old_digits;
            while(cur_digits >= new_digits) {
                res.push_back(int(cur % p[new_digits]));
                cur /= p[new_digits];
                cur_digits -= new_digits;
            }
        }
        res.push_back((int)cur);
        while(!res.empty() && res.back() == 0) {
            res.pop_back();
        }
        return res;
    }

    typedef vector<long long> vll;

    static vll karatsubaMultiply(const vll& a, const vll& b) {
        int n = a.size();
        vll res(n + n);
        if(n <= 32) {
            for(int i = 0; i < n; i++) {
                for(int j = 0; j < n; j++) {
                    res[i + j] += a[i] * b[j];
                }
            }
            return res;
        }

        int k = n >> 1;
        vll a1(a.begin(), a.begin() + k);
        vll a2(a.begin() + k, a.end());
        vll b1(b.begin(), b.begin() + k);
        vll b2(b.begin() + k, b.end());

        vll a1b1 = karatsubaMultiply(a1, b1);
        vll a2b2 = karatsubaMultiply(a2, b2);

        for(int i = 0; i < k; i++) {
            a2[i] += a1[i];
        }
        for(int i = 0; i < k; i++) {
            b2[i] += b1[i];
        }

        vll r = karatsubaMultiply(a2, b2);
        for(int i = 0; i < (int)a1b1.size(); i++) {
            r[i] -= a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            r[i] -= a2b2[i];
        }

        for(int i = 0; i < (int)r.size(); i++) {
            res[i + k] += r[i];
        }
        for(int i = 0; i < (int)a1b1.size(); i++) {
            res[i] += a1b1[i];
        }
        for(int i = 0; i < (int)a2b2.size(); i++) {
            res[i + n] += a2b2[i];
        }
        return res;
    }

    bigint operator*(const bigint& v) const {
        vector<int> a6 = convert_base(this->z, base_digits, 6);
        vector<int> b6 = convert_base(v.z, base_digits, 6);
        vll a(a6.begin(), a6.end());
        vll b(b6.begin(), b6.end());
        while(a.size() < b.size()) {
            a.push_back(0);
        }
        while(b.size() < a.size()) {
            b.push_back(0);
        }
        while(a.size() & (a.size() - 1)) {
            a.push_back(0), b.push_back(0);
        }
        vll c = karatsubaMultiply(a, b);
        bigint res;
        res.sign = sign * v.sign;
        for(int i = 0, carry = 0; i < (int)c.size(); i++) {
            long long cur = c[i] + carry;
            res.z.push_back((int)(cur % 1000000));
            carry = (int)(cur / 1000000);
        }
        res.z = convert_base(res.z, 6, base_digits);
        res.trim();
        return res;
    }
};

int n, k;
vector<array<int, 3>> triplets;

void read() {
    cin >> n >> k;
    triplets.resize(k);
    for(int i = 0; i < k; i++) {
        cin >> triplets[i][0] >> triplets[i][1] >> triplets[i][2];
    }
}

void solve() {
    // The number of ways to split 3n students into n unordered teams of three
    // with no constraints is (3n)! / (6^n * n!): order the students, then
    // divide out the 6 internal orderings of each team and the n! orderings of
    // the teams.
    //
    // The k forbidden triples are removed by inclusion-exclusion: for a chosen
    // subset S of forbidden triples that are pairwise disjoint (occupy 3*|S|
    // distinct students), the number of partitions that contain all of them is
    // the free count of the remaining 3n - 3|S| students, i.e.
    // (3n - cnt)! / (6^(n - cnt/3) * (n - cnt/3)!) with cnt = 3|S|, weighted by
    // (-1)^|S|. Subsets containing two triples sharing a student contribute 0.
    //
    // We aggregate the signs into coef[cnt] = sum of (-1)^|S| over all disjoint
    // subsets S with 3|S| = cnt, found by backtracking that marks used students
    // and skips conflicting triples. Then the answer is the sum over cnt of
    // coef[cnt] * (3n - cnt)! / (6^(n - cnt/3) * (n - cnt/3)!). All arithmetic
    // uses bigint since the factorials grow far beyond 64 bits.

    int total = 3 * n;
    vector<bigint> factorial(total + 1), power_6(total + 1);
    factorial[0] = 1;
    power_6[0] = 1;
    for(int i = 1; i <= total; i++) {
        factorial[i] = factorial[i - 1] * i;
        power_6[i] = power_6[i - 1] * 6;
    }

    vector<char> used(total, 0);
    vector<int64_t> coef(total + 1, 0);

    function<void(int, int, int64_t)> rec = [&](int pos, int cnt, int64_t p) {
        if(pos == k) {
            coef[cnt] += p;
            return;
        }

        rec(pos + 1, cnt, p);

        bool has_conflict = false;
        vector<int> to_undo;
        for(int j: triplets[pos]) {
            if(used[j - 1]) {
                has_conflict = true;
                break;
            }

            used[j - 1] = 1;
            to_undo.push_back(j - 1);
        }

        if(!has_conflict) {
            rec(pos + 1, cnt + 3, -p);
        }

        for(int j: to_undo) {
            used[j] = 0;
        }
    };

    rec(0, 0, 1);

    bigint ans = 0;
    for(int cnt = 0; cnt <= total; cnt++) {
        if(coef[cnt] != 0) {
            int rem = n - cnt / 3;
            bigint term = factorial[total - cnt] /
                          (power_6[rem] * factorial[rem]);
            ans += term * (int)coef[cnt];
        }
    }

    cout << ans << '\n';
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}


```

5. Python Solution

```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()
```
