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)

require 'set'

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

    if n > (b-1)
        until n % b == b-1
            n = n / b * (b-1) + (b-1)
        end
    end
    puts a*n/b - a*(n/b)
end

main

## イディオムなど
# until s.empty? do
#
## 配列のインデックス
# s[0,6] # 0番目から6文字
# s[0..6] # 0番目から6番目まで
# s[5..-1] # 5番目から最後まで
#
## 全探索
# [0, 1].repeated_permutation(n) do |combination| # bit全探索
# array.combination(i) do |selected| # array から要素が i 個である部分列を全部抽出
#
## 無向グラフの幅優先探索(BFS)
# to = Array.new(n).map{Array.new()} # ある頂点と繋がる頂点のリスト
# m.times do
#     a,b = gets.chomp.split(" ").map(&:to_i)
#     to[a].push b
#     to[b].push a
# end
# q = Queue.new # BFS は queue
# ds = Array.new(n, -1)
# pre = Array.new(n, -1)
#
## その他
# a1 = Marshal.load(Marshal.dump(a2)) # 深いコピー
# matrix = Array.new(n).map{Array.new()} # 二次元配列の初期化
#