Expected payoff of a zero-strike straddle on a normal

An underlying price at expiry is modelled as $S \sim N(0,1)$, and you hold a straddle struck at $K=0$, so the payoff is $|S-K|$.

Let $v$ be the expected payoff of this straddle. It turns out that $v^2 = \dfrac{a}{\pi}$ for some rational $a$. What is $a$?

$a = 2$. The payoff is $|S|$. By symmetry $\mathbb{E}|S| = 2\int_0^\infty \tfrac{x}{\sqrt{2\pi}}e^{-x^2/2}\,dx = \tfrac{2}{\sqrt{2\pi}} = \sqrt{2/\pi}$, so $v^2 = 2/\pi$. More generally $\mathbb{E}|S| = \sigma\sqrt{2/\pi}$ for $S\sim N(0,\sigma^2)$.