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)

def floor(a,b,x)
  (a*x/b).to_i- (x/b).to_i * a
end

nums=gets.strip.split(" ").map do |i|
  i.to_i
end
a=nums[0]
b=nums[1]
n=nums[2]
n_x=1
max = -9999999999
while n_x<=n 
  f=floor(a,b,n_x)
  if (max<f)
    max = f
  end

  n_x+=1
end
p max