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: Accepted)

#include<stdio.h>
long long a,b; 
long long f(long long x){return ((a*x)/b)-a*(x/b);}  
int main(){
  long long m,i,m1,m2,l=0,r;
  scanf("%lld %lld %lld",&a,&b,&r);
  r++;
  while(r-l>10){//printf("%lld %lld\n",l,r);
    m1=(l+l+r)/3LL;
    m2=(l+r+r)/3LL;
    //printf("%lld %lld\n",
    if(f(m1)<=f(m2))l=m1;
    else           r=m2+1;
  }
  m=0;
  for(i=l;i<r;i++){
    if(m<f(i))m=f(i);
  }
  printf("%lld\n",m);
  return 0;
}