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<math.h>
//using namespace std;
int main ()
{  long long a,b,n;
   scanf("%lld %lld %lld",&a,&b,&n);
   long long sum=0;
    long long i=1;
    if(n<=b)  {printf("%lld",(a*n)/b-a*(n/b));return 0;} 
   while(b*i<=n||b*i-1<=n)
   {    //if(i%b==0||i%b==b-1)
       long long ans=b*i;
       if(sum<(a*ans)/b-a*(ans/b))
	     sum=(a*ans)/b-a*(ans/b);
	     ans=b*i-1;
	     if(sum<(a*ans)/b-a*(ans/b))
	     sum=(a*ans)/b-a*(ans/b);
	     i++;
   }
   if(sum<(a*n)/b-a*(n/b)) sum=(a*n)/b-a*(n/b);
   printf("%lld",sum);
   return 0;
}