Implement a $3-$bit counter that counts in two different orders. Your circuit should have a Button named CLK$,$ a Button named RESET, and an input pin named G $($for Gray code$).$ It should have $3$ LED displays, named L$0,$ L$1,$ and L$2.$ The first thing the Test option does it ensure you have the correct inputs and outputs. The inputs and outputs are identical to those for the previous $3-$bit counter except for the addition of the extra pin named G$.$ The names are case sensitive$.$

When G is false, the counter is an up counter, counting in steps of $1.$ Each clock pulse increases the count by one: $0,1,2,3,4,5,6,7,0,$ etc.

When G is true, the counter is a gray code counter counting in gray code order: $0,1,3,2,6,7,5,4,$ then back to $0.$ In binary, that is $(000,001,011,010,110,111,101,100,$ then back to $000).$

Example: Suppose the current output is $101.$ If G$=1,$ the value after the next clock will be the next value in the gray code sequence, $100.$ If G$=0,$ the value after the next clock will be the next value sequentially, $110.$

the circuit should have a Button named CLK$,$ a Button named RESET, and an input pin named G$.$ It should have $3$ LED displays, named L$0,$ L$1,$ and L$2."$ It does not say it should have a clock. It has a clock button.

PLEASE DRAW$/$CREATE THE CIRCUIT DIAGRAM. This is to help me better understand this $3-$bit counter.