Localization and flatness

Let $M$ be a flat $S^{-1}A$-module. I want to show that $M$ is flat as an $A$-module.

$\newcommand{\tp}{\otimes}$ My "best" attempt is the following: Let $$0 \to N_1 \to N_2 \to N_3 \to 0$$ be exact (of $A$-modules). Then, $$0 \to S^{-1}N_1 \to S^{-1}N_2 \to S^{-1}N_3 \to 0$$ is exact, and as $M$ is flat, $$0 \to S^{-1}N_1 \tp_{S^{-1}A} M \to S^{-1}N_2 \tp_{S^{-1}A} M \to S^{-1}N_3 \tp_{S^{-1}A} M \to 0$$ is flat, and due to $S^{-1}N_i \tp_{S^{-1}A} M = S^{-1}N_i \tp_{S^{-1}A} S^{-1}M = S^{-1}(N_i \tp_A M)$ $$ 0 \to S^{-1}(N_1 \tp_A M) \to S^{-1}(N_2 \tp_A M) \to S^{-1}(N_3 \tp_A M) \to 0 $$ is exact. But I don't know why $$ 0 \to N_1 \tp_A M \to N_2 \tp_A M \to N_3 \tp_A M \to 0 $$ is exact, which would imply the flatness of $M$ as an $A$-module.

Just use that for any $A$-module $N$, one has the isomorphism $S^{-1}N\simeq N\otimes_AS^{-1}A$, so that $$(S^{-1}N)\otimes_{S^{-1}A}M\simeq (N\otimes_AS^{-1}A)\otimes_{S^{-1}A}M\simeq N\otimes_A M.$$