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<math.h>
int main()
{
	int a,b,n,temp,i,arr[100];
	scanf("%d%d%d",&a,&b,&n);
	for(i=0;i<=n;i++)
	{
		arr[i]=(floor((a*i)/b)-(a*(floor(i/b))));	
	}
	for(i=0;i<=n;i++)
	{
		if(arr[0]<arr[i])
		{
			temp=arr[0];
			arr[0]=arr[i];
			arr[i]=temp;
		}
	}
	printf("%d",arr[0]);
	return 0;
}