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: Time Limit Exceeded)

#include <stdio.h>
int main() {
  
long long int a, b, n;
scanf("%lld", &a);
scanf("%lld", &b);
scanf("%lld", &n);

long long int x = b - 1;
long long int t = 2;
long long int maxnum = -1;
long long int tmp = 0;

while(x < n){
  x = t * b - 1;
  tmp = a * x / b - x / b * a;
  if (maxnum < tmp){
    maxnum = tmp;
  }
  t++;
}
x = n;
tmp = a * x / b - x / b * a;
if (maxnum < tmp){
  maxnum = tmp;
}

printf("%lld", maxnum);

}