Consider an allocator that uses an implicit free list. The layout of each allocated and free memory block is as follows:

Header

Footer

$3$

$1$

$2$

$1$

Block Size $($bytes$)$

Block Size $($bytes$)$

Each memory block, either allocated or free, has a size that is a multiple of eight bytes. Thus, only the $29$ higher order bits in the header and footer are needed to record block size, which includes the header and footer. The usage of the remaining $3$ lower order bits is as follows:

$$ bit $0$ indicates the use of the current block: $1$ for allocated, $0$ for free.

$$ bit $1$ indicates the use of the previous adjacent block: $1$ for allocated, $0$ for free.

$$ bit $2$ is unused and is always set to be $0.$

Given the contents of the heap shown on the next page, show the new contents of the heap $($in the right table$)$ after a call to free$(0$x$400$b$010)$ is executed. Your answers should be given as hex values. Note that the address grows from bottom up$.$ Assume that the allocator uses immediate coalescing, that is$,$ adjacent free blocks are merged immediately each time a block is freed.

Give a short explanation of your answer: