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 <string.h>
#include <stdio.h>
#include <stdlib.h>
#include<math.h>
#define ll long long
long long min(long long a, long long b){
    if(a < b) return a;
    else return b;
}
long long max(long long a, long long b){
    if(a < b) return b;
    else return a;
}
int main(){
  ll int a,b,n,max=0,ans=0,tmp=0,point1,point2;
  scanf("%lld %lld %lld",&a,&b,&n);
  point1=(a*1)/b - a*(1/b);
  point2=(a*n)/b - a*(n/b);
  if(point1>=point2){
  for(ll int x=1;x<=n;x++){
    ans=(a*x)/b -a*(x/b);
  if(max>ans){
    printf("%lld",max);
    return 0;
  }
    max=ans;
  }
  }
  else{
    for(ll int x=n;x>=0;x--){
    ans=(a*x)/b -a*(x/b);
  if(max>ans){
    printf("%lld",max);
    return 0;
  }
    max=ans;
  }
  }
  return 0;
}