Problem:
Score : 400 points

Problem Statement

Given are integers A, B, and N.

Find the maximum possible value of floor(Ax/B) - A × floor(x/B) for a non-negative integer x not greater than N.

Here floor(t) denotes the greatest integer not greater than the real number t.

Constraints

1 ≤ A ≤ 10^{6}

1 ≤ B ≤ 10^{12}

1 ≤ N ≤ 10^{12}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B N

Output

Print the maximum possible value of floor(Ax/B) - A × floor(x/B) for a non-negative integer x not greater than N, as an integer.

Sample Input 1

5 7 4

Sample Output 1

2

When x=3, floor(Ax/B)-A×floor(x/B) = floor(15/7) - 5×floor(3/7) = 2. This is the maximum value possible.

Sample Input 2

11 10 9

Sample Output 2

9

Solution in Fortran (GNU Fortran 9.2.1) (judge verdict: Runtime Error)

program ksfloor
 implicit none
 integer :: n,answer,i
 integer,allocatable ::s(:)
 real :: a,b
 
 read*,a,b,n
 allocate(s(0:n))
 
 do i=0,n
  s(i)=aint(a*i/b)-a*aint(i/b)
 end do

 answer=maxval(s)
 print*,answer

end program ksfloor