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 C (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>  /*入出力*/
#include <stdlib.h> /*標準ライブラリ*/
#include <string.h> /*文字列*/
#include <math.h>   /*計算*/
#include <assert.h> /*確認*/
#include <ctype.h>  /*文字操作*/
#include <stdbool.h>/*bool型*/
#include <limits.h>/*限界*/
//---------------------------------------------------
#define rep(i, n) for(int  i = 0; i < (n); i++)
//#define max(a,b) ((a>b)?a:b)
#define ABS(a) (((a)>(0))?a:-(a))
#define LL long long
#define MOD 1000000007//10^9+7(素数)
//---------------------------------------------------

LL max(LL a, LL b){
  if (a > b) {
    return a;
  }else return b;
}


int main(void)
{
  double a,b,n;
  LL A,B;
  LL work1, work2;
  LL ans = 0;
  LL sum;
  scanf("%lf %lf %lf", &a, &b, &n);
  for (LL x = 1; x <= n; x++) {
    work1 = (LL)(a*((double)x)/b);
    work2 = (LL)(((double)x)/b);
    sum = work1 - ((LL)a) *work2;
    ans = max(ans,sum);
  }
  printf("%lld\n", ans);
}