1. Abridged Problem Statement  
Given integers A, α, β, γ, M and k, define a sequence x₀, x₁, … by  
  x₀ = A  
  xᵢ = (α·xᵢ₋₁² + β·xᵢ₋₁ + γ) mod M for i ≥ 1.  
Compute and output xₖ.

2. Detailed Editorial  
Problem essence: k can be as large as 10⁹, but M ≤ 1000, so the sequence values lie in [0, M−1] and must eventually repeat. We detect the first time we revisit a value (cycle entry) and then exploit the cycle structure to jump ahead.  

Step-by-step:  
1. If k = 0, answer is x₀ = A immediately.  
2. Otherwise, let x = A mod M, maintain an array `period[0…M−1]` initialized to −1. Also keep a counter `pos = 0`.  
3. While k > 0 and `period[x] == -1`:  
   a. Record `period[x] = pos++`.  
   b. Compute x = (α·x² + β·x + γ) mod M.  
   c. Decrement k.  
4. If k reaches 0 during this process, we have found xₖ; print x and stop.  
5. Otherwise, we have detected a cycle: let `cycle = pos - period[x]`. Now the remaining k steps all lie in a cycle of size `cycle`. We reduce with `k %= cycle`. Then simulate k more steps of x = f(x) mod M and print the result.  

Time complexity: O(M + cycle_length) ≤ O(M) ≈ O(1000). Memory: O(M).  

3. 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;
};

int a, alpha, bt, gamma, m, k;

void read() { cin >> a >> alpha >> bt >> gamma >> m >> k; }

int f(int x) { return (alpha * x * x + bt * x + gamma) % m; }

void solve() {
    // x_i = f(x_{i-1}) mod M, so the orbit starting at x_0 = A mod M enters a
    // cycle within at most M steps. We iterate f recording, in period[], the
    // first step index at which each residue is seen. If we exhaust the k
    // steps before revisiting a residue, the current value is the answer.
    // Otherwise we hit a repeat: the cycle length is the difference between the
    // current step count and when this residue first appeared, so we reduce the
    // remaining k modulo that length and finish by iterating f.

    if(k == 0) {
        cout << a << '\n';
        return;
    }

    vector<int> period(m, -1);
    int pos = 0, x = a % m;

    while(k > 0 && period[x] == -1) {
        period[x] = pos++;
        x = f(x);
        k--;
    }

    if(k == 0) {
        cout << x << '\n';
    } else {
        int cycle = pos - period[x];
        k %= cycle;
        while(k--) {
            x = f(x);
        }

        cout << x << '\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;
}
```

4. Python Solution

```python
import sys
def main():
    data = sys.stdin.read().split()
    # parse integers
    A, alpha, beta, gamma, M, k = map(int, data)

    # quick return if no steps
    if k == 0:
        print(A)
        return

    # function to compute next x
    def f(x):
        return (alpha * x * x + beta * x + gamma) % M

    # track first occurrence of each x in [0..M-1]
    first = [-1] * M
    x = A % M
    steps = 0

    # advance until either k runs out or x repeats
    while k > 0 and first[x] == -1:
        first[x] = steps
        steps += 1
        x = f(x)
        k -= 1

    # if we've completed k transitions, x is the answer
    if k == 0:
        print(x)
        return

    # otherwise, we hit a cycle
    cycle_start = first[x]
    cycle_len = steps - cycle_start

    # skip full cycles
    k %= cycle_len

    # do the leftover steps
    for _ in range(k):
        x = f(x)

    print(x)

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

5. Compressed Editorial  
We need xₖ for the quadratic recurrence mod M. Since M ≤ 1000, values repeat within ≤ M steps. Track first occurrence of each residue in an array; once a repeat is found, we identify the cycle length = current_step − first_occurrence[x]. Reduce remaining k by k mod cycle_length, then simulate the residue transitions for the leftover steps. Total time O(M).
