Calculating power of a Hypothesis Testing Problem based on Uniform distribution

Consider the problem of testing $H_0:a=0$ against $H_1:a=1/2$ based on a single observation X from U(a,a+1). The power of the test "Reject $H_0$ if $X>2/3$" is

(A)1/6 (B)5/6 (C)1/3 (D)2/3

My Steps:

Power of Test=P(reject $H_0$|$H_1$ is true)

c.d.f. of continuous Uniform distribution is given be $\frac{x-a}{b-a}$, where a and b are parameters of the given Uniform distribution, $U(a,b)$.

\begin{align} \text{Power of Test} & = P(\text{reject } H_0|H_1 \text{is true)} \\ & = P(X>2/3|a=1/2) \\ & = \frac{(2/3-1/2)}{1} \\ & = 1/6 \end{align}

No. It should be 5/6. Remember cdf of X is $P(X \leq x)$.

$P(X > 2/3|a = 1/2)$
$= 1 - P(X \leq 2/3|a = 1/2)$
$= 1 - (2/3 - 1/2)/1$
$= 5/6$