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For $K$-armed bandits with a unique best arm, the optimal sample complexities for both settings have been settled down, and they match up to logarithmic factors. This prompts an interesting research question about the generic, potentially structured BAI problems: is FB harder than FC or the other way around? In this paper, we show that FB is no harder than FC up to logarithmic factors. We do this constructively: we propose a novel algorithm called FC2FB (fixed confidence to fixed budget), which is a meta algorithm that takes in an FC algorithm $\\mathcal{A}$ and turn it into an FB algorithm. We prove that FC2FB enjoys a sample complexity that matches, up to logarithmic factors, that of the sample complexity of $\\mathcal{A}$. 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