Consider the following C code:

main $()\{$

int A$[2][2][3]$;

int i$,$j$,$k$,$x;

x$=0$;

for $($k$=0$; k$<3$; k$++)\{$

for $($j$=0$; j$<2$; j$++)\{$

for $($i$=0$; i$<2$; i$++)\{$

x$++$;

A$[$i$][$j$][$k$]=$ x;

$\}\}\}$

x$=-1$;

when row major is used

A$[0,0,0]=1$ A$[,,]=$ A$[,,]=$ A$[,,]=$ A$[,,]=$ A$[,,]=$ A$[,,]=$ A$[,,]=$ A$[,,]=$ A$[,,]=$ A$[,,]=$ A$[1,1,2]=12$

when column major is used

A$[0,0,0]=1$ A$[,,]=$ A$[,,]=$ A$[,,]=$ A$[,,]=$ A$[,,]=$ A$[,,]=$ A$[,,]=$ A$[,,]=$ A$[,,]=$ A$[,,]=$ A$[1,1,2]=12$

$\}$

Consider the consecutive blocks of memory given above on the right, where each block can keep an integer value. Assuming that A$[0,0,0]$ is stored in the first block, give the elements of the array A that will be stored in these consecutive blocks, together with their value, just before the statement x$=-1$ is executed.