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 <math.h>

int compare_int(const void *a, const void *b){
	return *(int*)a - *(int*)b;
}

int main(void){
	long double a,b;
	long long n;
	scanf("%Lf",&a);scanf("%Lf",&b);scanf("%lld",&n);
	long int f[n+1];
	for(int i=0;i<=n;i++){
		f[i] = (long int)floorl(a*i/b)-(long int)a*(long int)floorl(i/b);
	}

	qsort(f, n+1, sizeof(long int), compare_int);
	printf("%ld\n",f[n]);
	return 0;
}