Let $a$ and $b$ be two nonzero real numbers. If the coefficient of $x^5$ in the expansion of $\left(a x^2+\frac{70}{27 b x}\right)^4$ is equal to the coefficient of $x^{-5}$ in the expansion of $\left(\mathrm{ax}-\frac{1}{\mathrm{bx}^2}\right)^7$, then the value of $2 b$ is
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