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: Wrong Answer)

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
long long int floor_(float x){
 return (long long int)x/1;}

int main(){
 long long a,b,n;
 scanf("%lld%lld%lld",&a,&b,&n);
 
 if(n<b){
  printf("%lld\n",floor_(a*(float)n/(float)b));
  return 0;}
 
 if(n==b){
  printf("%lld\n",floor_(a*(float)(n-1)/(float)b));
  return 0;}
  
 else{
   long long x=n/b;
   x=x*b-1;
   printf("lld\n",floor_((float)a*x/(float)b)-a*floor_((float)x/(float)b));
  return 0;}

}