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 `first_occurrence[0…M−1]` initialized to −1. Also keep a counter `steps = 0`.  
3. While k > 0 and `first_occurrence[x] == -1`:  
   a. Record `first_occurrence[x] = steps++`.  
   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 `start = first_occurrence[x]`, `length = steps - start`. Now k remaining steps all lie in the cycle of size `length`. We can do `k %= length` to reduce. 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. Provided C++ Solution with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Read global parameters
int A, alpha, beta, gamma, M, k;

// Reads input values into the globals
void read() {
    cin >> A >> alpha >> beta >> gamma >> M >> k;
}

// Computes f(x) = (alpha*x^2 + beta*x + gamma) mod M
int f(int x) {
    return ( (long long)alpha * x * x + (long long)beta * x + gamma ) % M;
}

void solve() {
    // If k=0, answer is just A (no evolution)
    if(k == 0) {
        cout << A << '\n';
        return;
    }
    // period[x] will store at what step value x first appeared, or -1 if never seen
    vector<int> period(M, -1);
    int steps = 0;        // counts how many transitions we've done
    int x = A % M;        // current sequence value (always in [0, M-1])

    // Walk until we've done k steps or until we hit a repeated x
    while(k > 0 && period[x] == -1) {
        period[x] = steps;    // record we saw x at 'steps'
        steps++;
        x = f(x);             // move to next term
        k--;
    }

    // If we exhausted k, x now is x_k
    if(k == 0) {
        cout << x << '\n';
    } else {
        // Otherwise, we found a cycle: the first time we saw x was at period[x]
        int cycle_start = period[x];
        int cycle_length = steps - cycle_start;
        // We can skip full cycles
        k %= cycle_length;
        // Simulate the remaining k steps
        while(k--) {
            x = f(x);
        }
        cout << x << '\n';
    }
}

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

    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  
```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).