1. Abridged Problem Statement  
You generate a sequence {x₀, x₁, …} by  
• x₀ = 1  
• xᵢ = (A·xᵢ₋₁ + (xᵢ₋₁ mod B)) mod C  
Find the smallest index m such that x_m appears again later, and output the index of its second occurrence. If that second occurrence exceeds 2·10⁶, print –1.

2. Detailed Editorial  
We seek the first “repeatable” element in an infinite sequence defined by x₀=1 and x→f(x) where  
  f(x) = (A·x + (x mod B)) mod C.  
By definition, the first repeatable element M is the earliest xₘ for which there exists q>m with x_q = xₘ; we must output that q if q≤2·10⁶, otherwise –1.

Observation: Any sequence defined by iterating a deterministic function on a finite set must eventually enter a cycle:  
  1) A transient of length μ (possibly zero) before the cycle starts.  
  2) A cycle of length λ that repeats indefinitely.  
The first repeated value is x_μ (the first element of the cycle), and its next occurrence is at index μ+λ. Thus the answer is μ+λ, unless μ+λ>2·10⁶.

To compute μ and λ in O(μ+λ) time and O(1) memory, use Floyd’s “tortoise and hare” algorithm:  
  Phase 1: Find a meeting point inside the cycle.  
   – Initialize tortoise = f(1), hare = f(f(1)).  
   – Move tortoise one step, hare two steps, until they meet or until the step count exceeds a safe bound (>2·10⁶).  
  Phase 2: Find μ.  
   – Reset tortoise = x₀ = 1, keep hare at meeting point.  
   – Move both one step at a time; the index μ where they meet is the start of the cycle.  
  Phase 3: Find λ.  
   – From x_μ, move hare = f(tortoise) and count steps until it returns to tortoise; that count is λ.  
  Compute ans = μ + λ; if ans>2·10⁶, print –1, else print ans.

Complexity: Each function application f(x) is O(1). Phase 1 and 2 each take at most μ+λ steps, Phase 3 takes λ. Total O(μ+λ). As μ+λ≤C (and we abort above 2·10⁶), it’s efficient in 1 s.

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

// Overload operator<< for pair to ease debug printing.
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

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

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

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

// Global parameters
int64_t A, B, C;

// The transition function f(x) = (A*x + (x mod B)) mod C
int64_t f(int64_t x) {
    int64_t res = A;       // res = A
    res *= x;               // res = A * x
    res += x % B;           // res = A*x + (x mod B)
    res %= C;               // res = ( ... ) mod C
    return res;
}

// Read A, B, C from stdin
void read() {
    cin >> A >> B >> C;
}

// Apply Floyd's cycle-finding algorithm
int hare_tortoise() {
    // Step counter to prevent infinite loops beyond limit
    int t = 0;
    // Initialize tortoise = f(1), hare = f(f(1))
    int64_t tortoise = f(1);
    int64_t hare = f(f(1));

    // Phase 1: find a meeting point inside the cycle
    do {
        tortoise = f(tortoise);         // move tortoise 1 step
        hare = f(f(hare));              // move hare 2 steps
        t++;
        if (t > 2000042) {              // if we exceed ~2e6 steps
            return -1;                  // no repeat within limit
        }
    } while (tortoise != hare);

    // Phase 2: find mu = start index of cycle
    int mu = 0;
    tortoise = 1;                       // reset tortoise to x0
    while (tortoise != hare) {
        tortoise = f(tortoise);         // both move 1 step
        hare = f(hare);
        mu++;
    }

    // Phase 3: find lambda = cycle length
    int lambda = 1;
    hare = f(tortoise);                 // move hare one step from cycle start
    while (tortoise != hare) {
        hare = f(hare);
        lambda++;
    }

    // The second occurrence of x_mu is at index mu + lambda
    return mu + lambda;
}

// Solve one test case
void solve() {
    int ans = hare_tortoise();
    // If answer exceeds 2e6, print -1
    if (ans > 2000000) ans = -1;
    cout << ans << '\n';
}

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

    // Single test
    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
import sys
sys.setrecursionlimit(10**7)

def main():
    A, B, C = map(int, sys.stdin.readline().split())

    # Define the transition function f(x)
    def f(x):
        # (A*x + (x mod B)) mod C
        return (A * x + (x % B)) % C

    # Phase 1: detect a meeting in the cycle
    tortoise = f(1)          # moves 1 step
    hare = f(f(1))           # moves 2 steps
    steps = 0
    limit = 2_000_000 + 42
    while tortoise != hare and steps <= limit:
        tortoise = f(tortoise)
        hare = f(f(hare))
        steps += 1

    # If no meet within our limit, no repeat <=2e6
    if steps > limit:
        print(-1)
        return

    # Phase 2: find mu (start index of cycle)
    mu = 0
    tortoise = 1             # reset tortoise to x0
    while tortoise != hare:
        tortoise = f(tortoise)
        hare = f(hare)
        mu += 1

    # Phase 3: find lambda (cycle length)
    lam = 1
    hare = f(tortoise)
    while tortoise != hare:
        hare = f(hare)
        lam += 1

    ans = mu + lam
    # If the second occurrence is beyond 2e6, print -1
    print(ans if ans <= 2_000_000 else -1)

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

5. Compressed Editorial  
- The sequence x₀=1, xᵢ=f(xᵢ₋₁) must eventually cycle.  
- The first repeated value is at the cycle start index μ, and its next occurrence is at μ+λ, where λ is the cycle length.  
- Use Floyd’s tortoise-and-hare:  
  Phase 1: find a meeting point inside the cycle.  
  Phase 2: find μ by resetting one pointer to start and advancing both by one until they meet.  
  Phase 3: find λ by counting steps to return to the cycle start.  
- Answer = μ+λ if ≤2·10⁶, else –1.