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 (Clang 10.0.0) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
 
int main(void){
	long long A,B,N,ans,buff1,buff2,max;
	int i;
	char str[10000000],buf[10000000];
 
	fgets(str, sizeof(str), stdin);
	A=strtoll(strtok(str ," "),buf,10);
	B=strtoll(strtok(NULL," "),buf,10);
	N=strtoll(strtok(NULL," "),buf,10);

  max=0;
  
	for(i=0;i<=N;i++){
		buff1=A*i/B;
		buff2=i/B;
		buff2*=A;
		ans=buff1-buff2;
      	if(ans > max) max=ans;
	}
  
  	/*if(N>B){
      	ans=B;
    }else{
 		buff1=A*N/B;
		buff2=N/B;
		buff2*=A;
      	ans=buff1-buff2;
    }*/
  
	printf("%ld\n",max);

	return 0;
}