Smallest integers after N divisible by 2, 3, and 4 Give credit to whom credit is due.

Objective Given an integer N &gt; 0, out the smallest integers A, B, and C so that:


All of A, B, and C are strictly greater than N;
2 divides A;
3 divides B;
and 4 divides C.


This is a code-golf, so the shortest answer in bytes wins. You may use a language made/updated after this challenge, but it is strictly non-competing.

Test cases

N =&gt; A, B, C
1 =&gt; 2, 3, 4
4 =&gt; 6, 6, 8
43 =&gt; 44, 45, 44
123 =&gt; 124, 126, 124
420 =&gt; 422, 423, 424
31415 =&gt; 31416, 31416, 31416
1081177 =&gt; 1081178, 1081179, 1081180
Python 2, 32 bytes

lambda n:[n+2&amp;-2,n/3*3+3,n+4&amp;-4]

Bit arithmetic for 2 and 4, modular arithmetic for 3.
I found four 7-byte expressions for the next multiple of k above n but none shorter:
n-n%k+k
~n%k-~n
n/k*k+k
~n/k*-k

Any gives 34 bytes when copies for k=2,3,4, and 33 bytes if combined:
[n/2*2+2,n/3*3+3,n/4*4+4]
[n/k*k+k for k in 2,3,4]

But, 2 and 4 are powers of 2 that allow bit tricks to zero out the last 1 or 2 bytes.
n+2&amp;-2
n+4&amp;-4

This gives 6 bytes (instead of 7) for getting the next multiple, for 32 bytes overall, beating the for k in 2,3,4.
Unfortunately, the promising-looking n|1+1 and n|3+1 have the addition done first, so incrementing the output takes parentheses.