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 <ctype.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <stdbool.h>

#define loop(i,st,en) for(int i=st;i<en;i++)
#define loop0n(i,n) for(int i=0;i<n;i++)
#define min(a,b) ((a)>(b)?(b):(a))




int main() {
  unsigned long long int A=0,B=0,N=0,f1=0,f2=0;
  scanf("%I64lld %I64lld %I64lld",&A,&B,&N);
  f1=floor(A*(B-1)/B)-A*floor((B-1)/B);
  f2=floor(A*N/B)-A*floor(N/B);
  if((B-1)>N){
    printf("%I64lld\n",f2);
  }else{
    printf("%I64lld\n",min(f1,f2));
  }

  return 0;
}