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{|e|e.to_i}

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

def doit(a,b,n)
	if b>n then
		return check(a,b,n)
	else
		max = 0
		x = b-1
		while x<=n do
			c = check(a,b,x)
			if c>max then max = c end
			x += b
		end
		c = check(a,b,n)
		if c>max then max = c end
		return max
	end
end

print doit(a,b,n)

#5 7 20