<|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.

455. Sequence analysis
Time limit per test: 1 second(s)
Memory limit: 4096 kilobytes
input: standard
output: standard

Due to the slow 'mod' and 'div' operations with int64 type, all Delphi solutions for the problem 455 (Sequence analysis) run much slower than the same code written in C++ or Java. We do not guarantee that Delphi solution exists.


You are given a sequence of signed 64-bit integers defined as follows:
x0 = 1,
,
where mod is a remainder operator. All arithmetic operations are evaluated without overflow checking. Use standard "remainder" operator for programming languages (it differs from the mathematical version; for example  in programming, while  in mathematics). Use "long long" type in C++, "long" in Java and "int64" in Delphi to store xi and all other values.

Let's call a sequence element xp repeatable if it occurs later in the sequence — meaning that there exists such q, q > p, that xq = xp. The first repeatable element M of the sequence is such an element xm that xm is repeatable, and none of the xp where p < m are repeatable.

Given A, B and C, your task is to find the index of the second occurence of the first repeatable element M in the sequence if the index is less or equal to 2 · 106. Per definition, the first element of the sequence has index 0.

Input
The only line of input contains three signed 64-bit integers: A, B and C (B > 0, C > 0).

Output
Print a single integer  — the index of the second occurence of the first repeatable member if it is less or equal to 2 · 106. Print -1 if the index is more than 2 · 106.

Example(s)
sample input
sample output
2 2 9
4

sample input
sample output
2305843009213693951 1 9223372036854775807
5

sample input
sample output
-2 1 5
4



Note
In the first sample test the sequence starts with the following numbers: 1, 3, 7, 6, 3, 7. The first repeatable element is 3. The second occurence of 3 has index 4.

In the second sample test the sequence starts with the following numbers: 1, 2305843009213693951, -4611686018427387903, 6917529027641081855, 0, 0, 0. The first repeatable element is 0. The second occurence of 0 has index 5.

In the third sample test the sequence starts with the following numbers: 1, -2, 4, -3, 1, -2, 4. The first repeatable element is 1. The second occurence of 1 has index 4.

<|response|>
1. Abridged Problem Statement
You generate an infinite sequence {x₀, x₁, x₂, …} of signed 64-bit integers by:
  • x₀ = 1
  • xᵢ = (A·xᵢ₋₁ + (xᵢ₋₁ mod B)) mod C,   for i ≥ 1
Here B>0 and C>0. An element xₚ is called "repeatable" if it appears again later in the sequence (i.e., x_q = xₚ for some q>p). The first repeatable element M is the one at the smallest index m with that property. We must output the index of its second occurrence q (the smallest q>m with x_q = x_m), provided q ≤ 2·10⁶; otherwise print –1.

2. Key Observations
- Any sequence defined by iterating a deterministic function f on a finite set must eventually enter a cycle.
  - There is a "transient" (non-repeating prefix) of length μ ≥ 0, then a cycle of length λ ≥ 1 that repeats forever.
- The first repeated value is xₘ where m = μ (the first point where the cycle begins). Its next occurrence is at index μ+λ.
- Therefore the answer is μ+λ if μ+λ ≤ 2·10⁶, otherwise –1.
- Floyd's "tortoise and hare" algorithm finds μ and λ in O(μ+λ) time and O(1) extra space:
  1. Phase 1: move tortoise by f(x) and hare by f(f(x)) until they meet—guaranteed inside the cycle.
  2. Phase 2: reset tortoise to the start x₀; move both one step at a time until they meet again; that meeting index is μ.
  3. Phase 3: from that meeting point, move hare one step at a time until it returns; the count of steps is λ.

3. Full Solution Approach
1. Read A, B, C as signed 64-bit integers.
2. Define a function f(x):
     res = A*x
     res += x % B          (use the language's remainder operator)
     res %= C              (ensure result is in [0, C-1] if C>0)
     return res
3. Apply Floyd's algorithm:
   Phase 1 (find meeting point):
     tortoise = f(1)
     hare     = f(f(1))
     repeat up to a safe step bound (>2·10⁶) until tortoise == hare.
     If we exceed the bound, conclude the cycle's second occurrence index >2·10⁶ and print –1.
   Phase 2 (find μ):
     reset tortoise = 1, keep hare where it met.
     initialize mu = 0.
     while tortoise != hare:
       tortoise = f(tortoise)
       hare     = f(hare)
       mu++
   Phase 3 (find λ):
     initialize lambda = 1
     hare = f(tortoise)
     while tortoise != hare:
       hare = f(hare)
       lambda++
   The second occurrence index = mu + lambda.
4. If (mu + lambda) ≤ 2·10⁶, print that; otherwise print –1.

4. C++ Implementation
```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;
};

int64_t A, B, C;

int64_t f(int64_t x) {
    int64_t res = A;
    res *= x;
    res += x % B;
    res %= C;
    return res;
}

void read() { cin >> A >> B >> C; }

int hare_tortoise() {
    int t = 0;
    int64_t hare = f(f(1));
    int64_t tortoise = f(1);
    do {
        tortoise = f(tortoise);
        hare = f(f(hare));
        t++;
        if(t > 2000042) {
            return -1;
        }
    } while(tortoise != hare);

    int mu = 0;
    tortoise = 1;
    while(tortoise != hare) {
        tortoise = f(tortoise);
        hare = f(hare);
        mu++;
    }

    int lambda = 1;
    hare = f(tortoise);
    while(tortoise != hare) {
        hare = f(hare);
        lambda++;
    }

    return mu + lambda;
}

void solve() {
    // The sequence x_{i+1} = (A * x_i + x_i % B) % C is a functional graph
    // iteration f, so it eventually enters a cycle (rho shape). The first
    // repeatable element is the first one that lies on the cycle, i.e. the
    // entry point of the tail of length mu into the cycle of length lambda. Its
    // second occurrence is exactly mu + lambda steps from the start.
    //
    // We run Floyd's tortoise and hare: advance until they meet inside the
    // cycle, then find mu by walking both one step at a time from x_0 and the
    // meeting point until they coincide, and find lambda by walking one pointer
    // around the cycle. If no meeting happens within the search bound the index
    // exceeds 2 * 10^6, so we report -1.

    int ans = hare_tortoise();
    if(ans > 2000000) {
        ans = -1;
    }
    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 Implementation
```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()
```
