Write a Java application that asks for a binary number and outputs its decimal equivalent. Assume that the user will input valid binary numbers that are no more than 19 digit. For example 1111111111111111111 is a valid input but 11111111111111111111 is not a valid input Do not use Math.pow() to calculate exponents. Also you have to use loops instead of copy-pasting code. This is how the decimal equivalent of a binary number is calculated: (1011)2 = (1* 2) + (1*2') + (0*22) + (1 *23) in decimal So, the following is the pseudo-code of the algorithm that you must follow: Create a place to hold the decimal equivalent. Get a binary number. As long as the binary number is greater than 0 Add an appropriate value to the decimal equivalent (the appropriate value depends on the rightmost digit of the binary number) Get rid of the rightmost digit of the binary number o O