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 cal(int x){
  if(x == N + 1) return 0;

  int a=(int)(A*x/B);
  int b= (int)x/B;

  rem[x] = a - b*A;
  if(a - b*A == 0)  cal(x + 1);

  if(rem[x] >= rem[x-1])   result = rem[x];
  cal(x + 1);
}

int main(void){

  scanf("%d %d %d",&A,&B,&N);
  cal(1);
  printf("%d",result);
}