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: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
typedef long long ll;

int main()
{
	ll A,B,N;
	scanf("%lld %lld %lld", &A,&B,&N);
	ll i = N;
	ll x;
	ll n;
	ll sum;
	if (N >= B - 1)
		x = B - 1;
	else if (N < B)
		x = N;
	sum = (A*x/B) - A * (x/B);
	printf("%lld\n", sum);
	return (0);
}