from inline_snapshot import snapshot

from mirascope.llm import (
    AssistantMessage,
    Text,
    UserMessage,
)

test_snapshot = snapshot(
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                        Text(
                            text='''\
The problem asks us to calculate a "special sum" for a given integer `n`. The special sum is the sum of all integers `k` from 1 to `n` such that `k` is *not* a perfect square and `k` is *not* a perfect cube.

This problem can be efficiently solved using the Principle of Inclusion-Exclusion.

Let `S_total` be the sum of all integers from 1 to `n`.
Let `S_squares` be the sum of all perfect squares less than or equal to `n`.
Let `S_cubes` be the sum of all perfect cubes less than or equal to `n`.
Let `S_both` be the sum of all numbers that are *both* perfect squares *and* perfect cubes less than or equal to `n`.

The sum of numbers that are *either* perfect squares *or* perfect cubes (or both) is `S_squares + S_cubes - S_both`.
We want to find the sum of numbers that are *neither* perfect squares *nor* perfect cubes. This can be calculated as:
`Special Sum = S_total - (S_squares + S_cubes - S_both)`
`Special Sum = S_total - S_squares - S_cubes + S_both`

Let's break down each component:

1.  **`S_total` (Sum of integers from 1 to `n`):**
    This is a standard arithmetic series sum: `n * (n + 1) / 2`.

2.  **`S_squares` (Sum of perfect squares <= `n`):**
    These are `1^2, 2^2, ..., m^2` where `m` is the largest integer such that `m^2 <= n`. So, `m = floor(sqrt(n))`.
    The sum of the first `m` squares is given by the formula: `m * (m + 1) * (2*m + 1) / 6`.

3.  **`S_cubes` (Sum of perfect cubes <= `n`):**
    These are `1^3, 2^3, ..., p^3` where `p` is the largest integer such that `p^3 <= n`. So, `p = floor(cbrt(n))`.
    The sum of the first `p` cubes is given by the formula: `(p * (p + 1) / 2)^2`.

4.  **`S_both` (Sum of numbers that are both perfect squares and perfect cubes <= `n`):**
    A number `k` is both a perfect square (`a^2`) and a perfect cube (`b^3`) if and only if `k` is a perfect sixth power (`c^6`). This is because the exponents in its prime factorization must be multiples of both 2 and 3, meaning they must be multiples of `lcm(2, 3) = 6`.
    So, these are `1^6, 2^6, ..., q^6` where `q` is the largest integer such that `q^6 <= n`. So, `q = floor(n^(1/6))`.
    There isn't a simple closed-form formula for the sum of `i^6`. However, `q` will be relatively small even for very large `n`. For `n = 10^18`, `q = floor((10^18)^(1/6)) = floor(10^3) = 1000`. We can simply iterate from `i = 1` to `q` and sum `i^6`.

**Handling Large `n` and Floating Point Precision:**

*   `n` can be up to `10^18`, so Python's arbitrary-precision integers are necessary for sums.
*   Calculating `floor(n**(1/k))` (for `k=2, 3, 6`) requires care due to potential floating-point precision issues. A robust way is to estimate using `int(n**(1/k))` and then adjust the result with a small `while` loop to ensure it's the correct floor. Python 3.8+ offers `math.isqrt` for integer square roots, which is the most robust for `k=2`. For `k=3` and `k=6`, manual adjustment is suitable.

**Algorithm Steps:**

1.  Calculate `s_total = n * (n + 1) // 2`.
2.  Find `m = floor(sqrt(n))` using a robust method (e.g., `math.isqrt(n)` or the `get_kth_root_floor` helper function below).
3.  Calculate `s_squares = m * (m + 1) * (2 * m + 1) // 6`.
4.  Find `p = floor(cbrt(n))` using a robust method.
5.  Calculate `s_cubes = (p * (p + 1) // 2)**2`.
6.  Find `q = floor(n^(1/6))` using a robust method.
7.  Calculate `s_sixth_powers = sum(i**6 for i in range(1, q + 1))`.
8.  Return `s_total - s_squares - s_cubes + s_sixth_powers`.

**Robust `k`-th Root Floor Function:**

```python
def get_kth_root_floor(N, k):
    """
    Calculates floor(N^(1/k)) robustly for large N.
    """
    if N < 0:
        raise ValueError("N cannot be negative")
    if N == 0:
        return 0

    # Initial estimate using float power. This is usually close.
    root = int(N**(1/k))

    # Adjust the estimate to ensure it's the true floor.
    # Check if 'root' is too small
    while (root + 1)**k <= N:
        root += 1
    # Check if 'root' is too large
    while root**k > N:
        root -= 1
    return root
```

**Python Implementation:**

```python
import math

def get_kth_root_floor(N, k):
    """
    Calculates floor(N^(1/k)) robustly for large N.
    """
    if N < 0:
        raise ValueError("N cannot be negative")
    if N == 0:
        return 0
    \n\
    # For square root, math.isqrt is ideal in Python 3.8+
    if k == 2 and hasattr(math, 'isqrt'):
        return math.isqrt(N)

    # Initial estimate using float power. This is usually close.
    root = int(N**(1/k))

    # Adjust the estimate to ensure it's the true floor.
    # Check if 'root' is too small
    # Using 'root + 1' to avoid overflow issues if 'root**k' itself is already very large,
    # though for k=2,3,6 and N=10^18, this won't be an issue.
    while (root + 1)**k <= N:
        root += 1
    # Check if 'root' is too large
    while root**k > N:
        root -= 1
    return root

def special_sum(n: int) -> int:
    """
    Calculates the sum of k for all 1 <= k <= n such that k is not a perfect square
    and k is not a perfect cube.
    """
    if n < 1:
        return 0

    # 1. Total sum from 1 to n
    s_total = n * (n + 1) // 2

    # 2. Sum of perfect squares <= n
    m = get_kth_root_floor(n, 2)
    s_squares = m * (m + 1) * (2 * m + 1) // 6

    # 3. Sum of perfect cubes <= n
    p = get_kth_root_floor(n, 3)
    s_cubes = (p * (p + 1) // 2)**2

    # 4. Sum of numbers that are both perfect squares AND perfect cubes (i.e., perfect sixth powers) <= n
    q = get_kth_root_floor(n, 6)
    s_sixth_powers = 0
    for i in range(1, q + 1):
        s_sixth_powers += i**6
    \n\
    # Apply the Inclusion-Exclusion Principle
    result = s_total - s_squares - s_cubes + s_sixth_powers
    return result

```\
'''
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                    provider_id="google",
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                    provider_model_name="gemini-2.5-flash",
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                                "inline_data": None,
                                "text": '''\
The problem asks us to calculate a "special sum" for a given integer `n`. The special sum is the sum of all integers `k` from 1 to `n` such that `k` is *not* a perfect square and `k` is *not* a perfect cube.

This problem can be efficiently solved using the Principle of Inclusion-Exclusion.

Let `S_total` be the sum of all integers from 1 to `n`.
Let `S_squares` be the sum of all perfect squares less than or equal to `n`.
Let `S_cubes` be the sum of all perfect cubes less than or equal to `n`.
Let `S_both` be the sum of all numbers that are *both* perfect squares *and* perfect cubes less than or equal to `n`.

The sum of numbers that are *either* perfect squares *or* perfect cubes (or both) is `S_squares + S_cubes - S_both`.
We want to find the sum of numbers that are *neither* perfect squares *nor* perfect cubes. This can be calculated as:
`Special Sum = S_total - (S_squares + S_cubes - S_both)`
`Special Sum = S_total - S_squares - S_cubes + S_both`

Let's break down each component:

1.  **`S_total` (Sum of integers from 1 to `n`):**
    This is a standard arithmetic series sum: `n * (n + 1) / 2`.

2.  **`S_squares` (Sum of perfect squares <= `n`):**
    These are `1^2, 2^2, ..., m^2` where `m` is the largest integer such that `m^2 <= n`. So, `m = floor(sqrt(n))`.
    The sum of the first `m` squares is given by the formula: `m * (m + 1) * (2*m + 1) / 6`.

3.  **`S_cubes` (Sum of perfect cubes <= `n`):**
    These are `1^3, 2^3, ..., p^3` where `p` is the largest integer such that `p^3 <= n`. So, `p = floor(cbrt(n))`.
    The sum of the first `p` cubes is given by the formula: `(p * (p + 1) / 2)^2`.

4.  **`S_both` (Sum of numbers that are both perfect squares and perfect cubes <= `n`):**
    A number `k` is both a perfect square (`a^2`) and a perfect cube (`b^3`) if and only if `k` is a perfect sixth power (`c^6`). This is because the exponents in its prime factorization must be multiples of both 2 and 3, meaning they must be multiples of `lcm(2, 3) = 6`.
    So, these are `1^6, 2^6, ..., q^6` where `q` is the largest integer such that `q^6 <= n`. So, `q = floor(n^(1/6))`.
    There isn't a simple closed-form formula for the sum of `i^6`. However, `q` will be relatively small even for very large `n`. For `n = 10^18`, `q = floor((10^18)^(1/6)) = floor(10^3) = 1000`. We can simply iterate from `i = 1` to `q` and sum `i^6`.

**Handling Large `n` and Floating Point Precision:**

*   `n` can be up to `10^18`, so Python's arbitrary-precision integers are necessary for sums.
*   Calculating `floor(n**(1/k))` (for `k=2, 3, 6`) requires care due to potential floating-point precision issues. A robust way is to estimate using `int(n**(1/k))` and then adjust the result with a small `while` loop to ensure it's the correct floor. Python 3.8+ offers `math.isqrt` for integer square roots, which is the most robust for `k=2`. For `k=3` and `k=6`, manual adjustment is suitable.

**Algorithm Steps:**

1.  Calculate `s_total = n * (n + 1) // 2`.
2.  Find `m = floor(sqrt(n))` using a robust method (e.g., `math.isqrt(n)` or the `get_kth_root_floor` helper function below).
3.  Calculate `s_squares = m * (m + 1) * (2 * m + 1) // 6`.
4.  Find `p = floor(cbrt(n))` using a robust method.
5.  Calculate `s_cubes = (p * (p + 1) // 2)**2`.
6.  Find `q = floor(n^(1/6))` using a robust method.
7.  Calculate `s_sixth_powers = sum(i**6 for i in range(1, q + 1))`.
8.  Return `s_total - s_squares - s_cubes + s_sixth_powers`.

**Robust `k`-th Root Floor Function:**

```python
def get_kth_root_floor(N, k):
    """
    Calculates floor(N^(1/k)) robustly for large N.
    """
    if N < 0:
        raise ValueError("N cannot be negative")
    if N == 0:
        return 0

    # Initial estimate using float power. This is usually close.
    root = int(N**(1/k))

    # Adjust the estimate to ensure it's the true floor.
    # Check if 'root' is too small
    while (root + 1)**k <= N:
        root += 1
    # Check if 'root' is too large
    while root**k > N:
        root -= 1
    return root
```

**Python Implementation:**

```python
import math

def get_kth_root_floor(N, k):
    """
    Calculates floor(N^(1/k)) robustly for large N.
    """
    if N < 0:
        raise ValueError("N cannot be negative")
    if N == 0:
        return 0
    \n\
    # For square root, math.isqrt is ideal in Python 3.8+
    if k == 2 and hasattr(math, 'isqrt'):
        return math.isqrt(N)

    # Initial estimate using float power. This is usually close.
    root = int(N**(1/k))

    # Adjust the estimate to ensure it's the true floor.
    # Check if 'root' is too small
    # Using 'root + 1' to avoid overflow issues if 'root**k' itself is already very large,
    # though for k=2,3,6 and N=10^18, this won't be an issue.
    while (root + 1)**k <= N:
        root += 1
    # Check if 'root' is too large
    while root**k > N:
        root -= 1
    return root

def special_sum(n: int) -> int:
    """
    Calculates the sum of k for all 1 <= k <= n such that k is not a perfect square
    and k is not a perfect cube.
    """
    if n < 1:
        return 0

    # 1. Total sum from 1 to n
    s_total = n * (n + 1) // 2

    # 2. Sum of perfect squares <= n
    m = get_kth_root_floor(n, 2)
    s_squares = m * (m + 1) * (2 * m + 1) // 6

    # 3. Sum of perfect cubes <= n
    p = get_kth_root_floor(n, 3)
    s_cubes = (p * (p + 1) // 2)**2

    # 4. Sum of numbers that are both perfect squares AND perfect cubes (i.e., perfect sixth powers) <= n
    q = get_kth_root_floor(n, 6)
    s_sixth_powers = 0
    for i in range(1, q + 1):
        s_sixth_powers += i**6
    \n\
    # Apply the Inclusion-Exclusion Principle
    result = s_total - s_squares - s_cubes + s_sixth_powers
    return result

```\
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