How is this possible?

Let \(f(n)\) be the smallest divisor of \(n\) which is greater than or equal to \(\sqrt{n}\).

\[\large \lim_{m \to \infty} \frac{\ln(m)}{m^2} \sum\limits_{n=1}^{m}f(n)\]

If the value of the limit above is equal to \(\dfrac{\pi^A}{B}\) for integers \(A\) and \(B\), find \(A+B\).

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