<|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.

181. X-Sequence
time limit per test: 0.25 sec.
memory limit per test: 4096 KB
input: standard
output: standard



Let {xi} be the infinite sequence of integers:
1) x0 = A;
2) xi = (alpha * xi-1^2 + beta * xi-1 + gamma) mod M, for i >= 1.
Your task is to find xk if you know A, alpha, beta, gamma, M and k.

Input
Given A (1 <= A <= 10000), alpha (0 <= alpha <= 100), beta (0 <= beta <= 100), gamma (0 <= gamma <= 100), M (1 <= M <= 1000), k (0 <= k <= 10^9). All numbers are integer.

Output
Write xk.

Sample test(s)

Input
1 1 1 1 10 1

Output
3
Author:	Michael R. Mirzayanov
Resource:	Saratov ST team Spring Contest #1
Date:	18.05.2003

<|response|>
1. Abridged Problem Statement  
Given nonnegative integers A, α, β, γ, M and a (possibly large) index k, define the sequence  
  x₀ = A  
  xᵢ = (α·xᵢ₋₁² + β·xᵢ₋₁ + γ) mod M  for i ≥ 1.  
Compute and output xₖ.

2. Key Observations  
- Since every xᵢ is taken modulo M, there are only M possible distinct values: 0,1,…,M−1.  
- By the pigeonhole principle, the sequence must eventually revisit a previously seen value, forming a cycle.  
- If we record for each residue the step at which it first appears, then when we see it again we can determine:  
  • the start index of the cycle (where it first appeared)  
  • the cycle length = (current step) − (first occurrence step)  
- Once a cycle is detected, we can reduce the remaining number of steps k by taking k mod (cycle length), then simulate only those remaining steps.

3. Full Solution Approach  
a. Handle the trivial case k = 0: output A immediately.  
b. Work with x = A mod M (all further terms are in [0, M−1]).  
c. Create an array `first_occurrence` of size M, initialize all entries to –1.  
d. Maintain a step counter `steps = 0`.  
e. While k > 0 and the current x has not been seen before (`first_occurrence[x] == -1`):  
   1. Record `first_occurrence[x] = steps`.  
   2. Compute the next term via  
        x = (α·x² + β·x + γ) mod M  
   3. Increment `steps` and decrement `k`.  
f. If k reaches 0 in that loop, x is already xₖ. Output it.  
g. Otherwise, we have encountered a repeated value x. Let  
     cycle_start = first_occurrence[x]  
     cycle_length = steps − cycle_start  
   We still have k > 0 steps to go, but they lie entirely within the cycle. So set  
     k = k mod cycle_length  
   Then perform k further updates of x = f(x). The resulting x is xₖ. Output it.

Time complexity: O(M) to detect the cycle plus up to O(cycle_length) ≤ O(M) to finish. Since M ≤ 1000, this is very fast.  

4. C++ Implementation with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Globals to hold input parameters
int A, alpha, beta, gamma_val, M;
long long k;

// Read all inputs
void read_input() {
    cin >> A >> alpha >> beta >> gamma_val >> M >> k;
}

// Compute f(x) = (alpha*x^2 + beta*x + gamma) mod M
int next_value(int x) {
    // Use long long to avoid overflow in x*x
    long long t = (long long)alpha * x * x
                + (long long)beta * x
                + gamma_val;
    return int(t % M);
}

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

    read_input();

    // If no steps requested, answer is x0 = A
    if (k == 0) {
        cout << A << "\n";
        return 0;
    }

    // Array to record the first step at which each residue appeared
    vector<long long> first_occurrence(M, -1);

    // Start from x0 = A mod M
    int x = A % M;
    long long steps = 0;

    // Simulate until we run out of k or we revisit a residue
    while (k > 0 && first_occurrence[x] == -1) {
        first_occurrence[x] = steps;
        steps++;
        x = next_value(x);
        k--;
    }

    // If we exhausted the steps exactly, x is x_k
    if (k == 0) {
        cout << x << "\n";
        return 0;
    }

    // Otherwise, we've found a cycle beginning at first_occurrence[x]
    long long cycle_start = first_occurrence[x];
    long long cycle_length = steps - cycle_start;

    // Skip over as many full cycles as possible
    k %= cycle_length;

    // Perform the remaining k steps inside the cycle
    while (k-- > 0) {
        x = next_value(x);
    }

    cout << x << "\n";
    return 0;
}
```

5. Python Implementation with Detailed Comments  
```python
import sys

def main():
    data = sys.stdin.read().split()
    A, alpha, beta, gamma, M, k = map(int, data)

    # If k == 0, answer is simply A
    if k == 0:
        print(A)
        return

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

    # Track the first step at which each residue appears
    first_occurrence = [-1] * M

    # Start from x0 = A mod M
    x = A % M
    steps = 0

    # Walk until we either finish k steps or see a repeat
    while k > 0 and first_occurrence[x] == -1:
        first_occurrence[x] = steps
        steps += 1
        x = f(x)
        k -= 1

    # If we've used up all steps, print the current x
    if k == 0:
        print(x)
        return

    # Otherwise, a cycle has been detected at residue x
    cycle_start = first_occurrence[x]
    cycle_length = steps - cycle_start

    # Reduce k modulo the cycle length
    k %= cycle_length

    # Simulate the remaining k steps within the cycle
    for _ in range(k):
        x = f(x)

    print(x)

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