Hadamard in the ± basis

Let $\lvert\pm\rangle=\tfrac1{\sqrt2}(\lvert0\rangle\pm\lvert1\rangle)$ and $H=\tfrac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$ in the computational basis. Enter the $2\times2$ matrix of $H$ in the ordered basis $(\lvert+\rangle,\lvert-\rangle)$, whose entry in row $a$ and column $b$ is $\langle a\rvert H\lvert b\rangle$ for $a,b\in\{+,-\}$. Entries are accepted within $10^{-3}$.