The Clock Hands of Doom You are chained to a chair. Underneath you is a huge volcano. A 12-hour clock next to you ticks ominously, and you see that it has wires leading from the back up to a chain, which will drop you into the center of the earth. Taped to the clock is a note:


  Each clock hand has an electrode. When both clock hands are in the same position, the power flows and you die. That is, unless you can tell me the exact time that this will occur, to the nearest minute.


You have a computer that knows every programming language. You need to create the shortest (this is code-golf, and standard loopholes are prohibited) program you can, and tell the evil scientist what the time will be. Your program should take input (in any method), consisting of the hour and minute. It should return the next hour and minute (in any method) that this occurs.

According to the OEIS page, the eleven overlap times are:

00:00:00 plus 0/11 s, 01:05:27 plus 3/11 s,
02:10:54 plus 6/11 s, 03:16:21 plus 9/11 s,
04:21:49 plus 1/11 s, 05:27:16 plus 4/11 s,
06:32:43 plus 7/11 s, 07:38:10 plus 10/11 s,
08:43:38 plus 2/11 s, 09:49:05 plus 5/11 s,
10:54:32 plus 8/11 s.


The next time would be 12:00:00. The seconds and their fractional parts are not needed for this challenge. Simply round to the nearest minute.

Test cases:

0:00 (Or 12:00) &gt; 1:05
1:00 &gt; 1:05
11:56 &gt; 12:00 (Or 0:00)
6:45 &gt; 7:38
5:00 &gt; 5:27
6:30 &gt; 6:33 (round up)


The program can be a function, or full program. I do not care if you choose 0:00 or 12:00, and both are acceptable. Good luck!
JavaScript (Node.js), 54 47 bytes (round to the nearest)

-7 bytes. Thanks @user202729



a=&gt;b=&gt;[(a+=b&gt;5.46*a)+a/11|0,a%12*65.46%60+.5|0]


Try it online!



JavaScript (Node.js), 40 33 44 bytes (rounds towards 0)

-3 bytes thanks to @Arnauld

-4 bytes thanks to @Kevin Cruijssen



a=&gt;b=&gt;[(a+=b&gt;5.46*a)+a/11|0,a%12*65.46%60|0]


Explanation

a=&gt;b=&gt;[(a+=b&gt;5.46*a)+a/11|0,a%12*65.46%60|0]    Full Code
a                                               Hours
   b                                            Minutes
    =&gt;[                    ,               ]    return array with
       (a+=        )                            add to the current hour
           b&gt;5.46*a                             1 if the minute's clock hand has 
                                                passed the hour's clock hand. Here we use
                                                equation 60*a/11 which is the same as 5.46*a
                    +a/11                       and add 1 when hour is 11
                         |0                     floor the result

                            a%12*65.46%60|0     Here we do equation ((720/11)*a) (mod 60)
                            a%12                In case of hour 12 we take 0
                                *65.46          multiply hour by 720/11 which can be shortened to
                                                65.46 to save 1 byte.
                                      %60       mod 60
                                         |0     floor the result


Side note: I'm pretty sure this can be golf down by someone with more knowledge at math. I barely know how to sum and multiply

Try it online!