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: Time Limit Exceeded)

a, b, n = gets.chomp.split(" ").map(&:to_i);

max = 0
x =0
while (x < n) do
  if ((a*(x+100))/b)-a*((x+100)/b) >= max && x +100 < n
    max = [((a*(x+100))/b)-a*((x+100)/b), max].max
    x += 100
  elsif ((a*(x+10))/b)-a*((x+10)/b) >= max && x +10 < n
    max = [((a*(x+10))/b)-a*((x+10)/b), max].max
    x += 10
  elsif ((a*(x+1))/b)-a*((x+1)/b) >= max 
    max = [((a*(x+1))/b)-a*((x+1)/b), max].max
    x += 1
  else
    break if ((a*x)/b)-a*(x/b) == 0
  end
end

puts max