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 Julia (1.4.0) (judge verdict: Wrong Answer)

function main()
    a, b, n = parse.(UInt128, split(readline())) 
    before = f(one(Float64), a, b)
    hist = Float64[before]
    local res
    for x in 2:n
        y = f(Float64(x), a, b)
        push!(hist, y)
        if before == y
            break
        end
    end
    println(UInt128(maximum(hist)))
end

f(x,a,b)::Float64 = floor(a*x/b) - a*floor(x/b)

main()