Write a C program that prompts the user to enter the hours, minutes and seconds of a time in the day, t$1,$ say, hh$1,$ mm$1$ and ss$1,$ respectively. Then, it prompts the user to enter and for a later time in the day, t$2.$ The program computes the difference between the two time instants and prints the result in the form $$hh : mm : ss $.$ Assume the value of the hours is entered using the $24-$ hour notation $($rather than $12-$hour notation followed by am or pm$).$ For example, if the user enters t$2=15$ : $43$ : $27$ and t$1=06$ : $58$ : $49,$ the program outputs: The time difference t$2$ t$1$ is $08$ : $44$ :$38.$ Hint: Since t$2=$ hh$2$:mm$2$:ss$2$ and t$1=$ hh$1$:mm$1$:ss$1.$ The time difference is t$2$t$1=$hh:mm:ss where ss $=$ ss$2$ ss$1,$ mm $=$ mm$2$ mm$1$ and hh $=$ hh$2$ hh$1.$ If ss$1$ is greater than ss$2,$ the difference is negative. You want to fix that by borrowing one minute from mm$2($worth $60$ seconds to make ss positive. A similar argument applies when mm$1$ is larger than mm$2.$