The keys $14,18,13,42,3,23,25$ and $51$ are inserted into an initially empty hash table of length $11$ using hash function $h(k)=k$ mod $11$ and linear probing for collision resolution. What is the resultant hash table?

$\backslash$table$[[0,1,2,3,4,5,6,7,8,9,10,],[$Table A$,,,,,,,,,,,],[,23,13,14,,3,,18,25,42,51,]]$

$\backslash$table$[[0,1,2,3,4,5,6,7,8,9,10,$Table B$],[3,23,13,14,25,51,,18,,42,]]$

$\backslash$table$[[0,1,2,3,4,5,6,7,8,9,10,$Table C$],[,23,13,14,3,25,,18,51,42,]]$

$\backslash$table$[[0,1,2,3,4,5,6,7,8,9,10,$Table D$],[,14,13,23,,3,,18,25,42,51]]$

Select one:

a$.$ Table A

b$.$ Table B

c$.$ Table C

d$.$ Table D