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Given the expression, (2+1)x3 [Note that the expression use x for multiplication instead of * so please reflect this in your answer] convert it into PRN postfix by showing your working in the given table using the algorithm below. Initialize an empty stack of operators get next token in infix expression YES NO end of infix expression switch( token operand Push it onto the stack display it Pop and display stack items until the stack is empty Pop and display stack element until a left) is encountered, but don't display) If stack is empty or token has higher precedence than top stack element, then push token onto stack, otherwise, pop and display top stack element, then repeat the comparison of token with new top stack item terminate The initial values are given to you. Answer format: Your typed-in answers MUST NOT have any white space e.g. ABx and x+ is typed in without any white space. If the answer is empty, type empty in all lowercase. Token (2+1)x3 Output/Postfix empty Stack empty Reason token is (, push onto stack empty - empty Answer= Token (2+1)x3 Output/Postfix empty Stack empty Reason token is ( push onto stack empty no input remain, pop all token is operator and precedence > than top element in stack or stack is empty, push token into stack token is ) so pop all until (is found and discard both parentheses token is (, push onto stack token is operand, put token to postfix empty Answer= Given the expression, (2+1)x3 [Note that the expression use x for multiplication instead of * so please reflect this in your answer] convert it into PRN postfix by showing your working in the given table using the algorithm below. Initialize an empty stack of operators get next token in infix expression YES NO end of infix expression switch( token operand Push it onto the stack display it Pop and display stack items until the stack is empty Pop and display stack element until a left) is encountered, but don't display) If stack is empty or token has higher precedence than top stack element, then push token onto stack, otherwise, pop and display top stack element, then repeat the comparison of token with new top stack item terminate The initial values are given to you. Answer format: Your typed-in answers MUST NOT have any white space e.g. ABx and x+ is typed in without any white space. If the answer is empty, type empty in all lowercase. Token (2+1)x3 Output/Postfix empty Stack empty Reason token is (, push onto stack empty - empty Answer= Token (2+1)x3 Output/Postfix empty Stack empty Reason token is ( push onto stack empty no input remain, pop all token is operator and precedence > than top element in stack or stack is empty, push token into stack token is ) so pop all until (is found and discard both parentheses token is (, push onto stack token is operand, put token to postfix empty Answer=