Counting Solutions of a Reciprocal Equation

Given a positive integer n, consider the equation:

$$\frac{1}{x}+\frac{1}{y}=\frac{1}{n!}$$

where n! denotes the factorial of n. It can be shown by algebraic manipulation that the equation is equivalent to:

$$ (x-n!)(y-n!)=(n!)^2 $$

The number of positive integer solutions (x,y) is equal to the number of positive divisors of \((n!)^2\). Your task is to compute that number modulo \(10^9+7\).

Note: You may use the fact that if the prime factorization of \(n!\) is \(n! = \prod_{p \le n} p^{e_p}\), then the number of positive divisors of \((n!)^2\) is \(\prod_{p \le n} (2e_p+1)\).

inputFormat

The input consists of a single integer n (n ≥ 1).

outputFormat

Output a single integer representing the number of solution pairs \((x,y)\) modulo \(10^9+7\).

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