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Intuition\n Describe your first thoughts on how to solve this problem. \nThe goal is to count the number of ways to arrange numbers from 1 to n such that prime 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal is to count the number of ways to arrange numbers from 1 to $$n$$ such that prime numbers occupy prime indices. This problem is interesting because it combines elements of number theory (identifying prime numbers) and combinatorics (counting permutations).\n\nThe key insight is to recognize that the placement of prime numbers is independent of the specific prime indices. Instead, we only need to count the number of prime and non-prime numbers up to $$n$$, and then determine how many ways we can arrange these counts. By calculating the factorial of the number of primes and the factorial of the number of non-primes, we account for all possible arrangements.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. **Prime Checking:** We implement an efficient prime-checking function. For each number up to $$n$$, we check if it is prime. This is done by first eliminating small prime factors (2 and 3) and then checking for larger prime factors up to the square root of the number.\n\n2. **Counting Primes:** We iterate through all numbers from 1 to $$n$$, using the prime-checking function to count how many are prime.\n\n3. **Factorial Calculation:** We implement a function to calculate the factorial of a number modulo $$10^9 + 7$$. This function will be used to compute the factorial of the number of primes and non-primes.\n\n4. **Calculate Result:** We multiply the factorial of the number of primes by the factorial of the number of non-primes and take the result modulo $$10^9 + 7$$. This gives us the total number of valid permutations.\n\n# Complexity\nTime complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n  - Prime checking for each number from 1 to $$n$$ takes $$O(n \\sqrt{n})$$.\n  - Calculating factorials takes $$O(n)$$.\n  - Overall time complexity is $$O(n \\sqrt{n})$$.\n\nSpace complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n  - The algorithm uses $$O(1)$$ extra space for variables and results.\n\n# Code\n```python\nclass Solution:\n    def numPrimeArrangements(self, n: int) -> int:\n        MOD = 10**9 + 7\n        \n        def is_prime(num):\n            """\n            # Alternative:\n            for i in range(2, int(n**0.5) + 1):\n                if n % i == 0: return False\n            return True \n            """\n            if num <= 1:\n                return False\n            if num <= 3:\n                return True\n            if num % 2 == 0 or num % 3 == 0:\n                return False\n            i = 5\n            while i * i <= num:\n                if num % i == 0 or num % (i + 2) == 0:\n                    return False\n                i += 6 # Skip even numbers and multiples of 3\n            return True\n\n        def factorial(num):\n            f = 1\n            for i in range(2, num + 1):\n                f = f * i \n            return f\n        \n        primes_count = sum(is_prime(x) for x in range(1, n + 1))\n        non_primes_count = n - primes_count\n        return factorial(primes_count) * factorial(non_primes_count) % MOD\n```\nUPVOTE IF HELPFUL \uD83E\uDD19\nTHANKS
['Python3']