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 Ruby (2.7.1) (judge verdict: Accepted)

#require 'prime'
#require 'openssl'
MOD=10**9+7;  cnt=0; sum=0; prev=nil; can=true; h=Hash.new(0)
def gs() gets.chomp end
def gi() gets.chomp.to_i end
def gsmi() gets.chomp.split.map(&:to_i) end

def ar2(s1,s2,de=nil) Array.new(s1){Array.new(s2,de)} end
def ar3(s1,s2,s3,de=nil) Array.new(s1){Array.new(s2){Array.new(s3,de)}} end

def desc(ar) ar.sort!{|x,y| y<=>x} end
def min(a,b) a<=b ? a : b end
def max(a,b) a>=b ? a : b end
def sum(ar) ar.inject(:+).to_i end
def ceil(a,b) (a.to_f/b).ceil end

def C(a,b) b==0||a==b ? 1 : (b=a-b if a/2<b;(a-b+1..a).inject(:*)/(1..b).inject(:*)) end
def rui(ar) s=[0]; ar.size.times{|i| s<<s[-1]+ar[i]}; s end
def rev_rui(ar) s=[0]; ar.size.times.reverse_each{|i| s<<s[-1]+ar[i]}; s end

def puts_yesno(b) puts(b ? 'Yes' : 'No') end
def putsend(s) puts s; exit end
def pu(h) begin; require './../nemi_lib/debug.rb'; puts_val(h); rescue LoadError; end; end
########### ( ˘ω˘ )ｽﾔｧ… ###########
a,b,n=gsmi
x=min(b-1,n)
ans=(a.to_f*x/b).floor-a*(x.to_f/b).floor
puts ans