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: Runtime Error)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define BUFSIZE 1000000
int A,B,N;
long long result = 0;
long long rem[BUFSIZE]={0};


int int_sort( const void * a , const void * b ) {
  /* 引数はvoid*型と規定されているのでint型にcastする */
  if( *( int * )a < *( int * )b ) {
    return 1;
  }
  else
  if( *( int * )a > *( int * )b ) {
    return -1;
  }
  return 0;
}



int main(void){

  scanf("%d %d %d",&A,&B,&N);
  for(int i = 1;i < N + 1; i++){
    rem[i] = (A*i - (A*i % B))/B - A*(i - i % B)/B;
  }

  qsort((void *) rem, N + 1 ,sizeof(rem[0]) , int_sort);


printf("%d",rem[0]);

}