Counting Child Numbers

A positive integer \( n \) is called a Child Number if \( n^{61} \) divides \( 2024! \) evenly, i.e. \( 2024!/n^{61} \) leaves no remainder. Here, \( 2024! \) denotes the factorial of 2024: \(1 \times 2 \times \cdots \times 2024\).

In other words, if the prime factorization of \( n \) is given by \( n = \prod_{p} p^{a_p} \), then \( n \) is a Child Number if and only if for every prime \( p \) the inequality

[ 61,a_p \leq \sum_{i \geq 1} \left\lfloor \frac{2024}{p^i} \right\rfloor ]

holds. The number of choices for the exponent \( a_p \) for a given prime \( p \) is \( \left\lfloor \frac{v_p(2024!)}{61} \right\rfloor + 1 \) where \( v_p(2024!) = \sum_{i \geq 1}\!\left\lfloor \frac{2024}{p^i} \right\rfloor \).

Your task is to compute the total number of Child Numbers in the interval \([1, \infty)\). The final answer is given by the product over all primes \( p \) (that contribute non-trivially) of \( \left( \left\lfloor \frac{v_p(2024!)}{61} \right\rfloor + 1 \right) \).

inputFormat

This problem does not require any input.

outputFormat

Output a single integer representing the total number of Child Numbers. For the given problem, the expected output is 17978112.

sample

N/A

17978112