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In the case that $K$ admits some $S^1 \\times S^2$ surgery, $K$ is Floer simple, that is, the rank of $\\hat{HFK}(Y,K)$ is equal to the order of $H_1(Y)$. 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We prove that $K$ is rationally fibered, that is, the knot complement admits a fibration over $S^1$. As part of the proof, we show that if $K\\subset Y$ has a Dehn surgery to $S^1 \\times S^2$, then $K$ is rationally fibered. In the case that $K$ admits some $S^1 \\times S^2$ surgery, $K$ is Floer simple, that is, the rank of $\\hat{HFK}(Y,K)$ is equal to the order of $H_1(Y)$. 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