Determine the resulting stack which was empty at the start w

Determine the resulting stack, which was empty at the start while showing all steps: s - push(5); s - push(6); right = s.pop(); left = s.pop(); s - push(left - right); s - push(4); s - push(3); right = s.pop(); left = s.pop(); s - push(left * right); left = s.pop(); right = s.pop(); s - push(left - right);

Solution

let stack,s which is empty in start
push=insert element at the top of the stack
pop=remove element from the top of the stack

steps Result
s.push(5); s= 5
s.push(6); s=6,5
right=s.pop(); right=6,s=5
left=s.pop(); left=5,s is empty   
s.push(left - right); s=-1
s.push(4); s=4,-1
s.push(3); s=3,4,-1
right=s.pop(); right=3,s=4,-1
left=s.pop(); left=4,s=-1
s.push(left*right); s=12,-1
left=s.pop(); left=12,s=-1
right=s.pop(); right=-1,s is empty
s.push(left - right); s=13

Final answer:Resulting stack will have only one value i.e 13.

Please let me know in case of any question,thanks.