One day interesting work was assigned to the chef. The task is to do n unit of work before steeping, chef is already tired, so he decided to do v unit of work and take some energy drink, then he will do v$/$k $($the integral part when we divide v by k$)$ unit of work, again take some energy drink, then he will do vi$($k$^2)$ unit work and so on; vi$($k$^3)$v$($k$^)$v$/$k$5$v$/($k$^6)$ chef will sleep when the value of vi$($k$^$p$)$ is equal to $0.$ Help them to calculate the minimum value of v so that he will do at least in units of work before sleepingContest ends in $13$ hoursSubmissions: $102$Max Score: $30$Difficulty: MediumInput FormatRate This ChallengeThe input consists of two integers n and kConstraints$1$ sms $109\mathrm{.}2$sks$10$MoreOutput FormatPrint the only integer$-$the minimum value of v that lets chef will do work in one nightSampie Input D$2\mathrm{.}18$