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>


int main(void){

        long long int a,b,n;
        scanf("%lld%lld%lld",&a,&b,&n);
        long long int ans[n];

        long long int A[n],B[n];


        for(long long int i=0;i<=n;i++){
                A[i] = a*i/b;
                B[i] = i/b;
                ans[i] = A[i] - a * B[i];
        }   

        for(long long int i=0;i<=n;i++){
                for(long long int j=0;j<=n-i-1;j++){
                        if(ans[j]<ans[j+1]){
                                long long int tmp = ans[j];
                                ans[j] = ans[j+1];
                                ans[j+1] = tmp;
                        }   
                }   
        }   

        printf("%lld\n",ans[0]);
        return 0;
}