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: Accepted)

/*************************************************************************
	> File Name: 1.cpp
	> Author: Knowledge_llz
	> Mail: 925538513@qq.com 
	> Blog: https://blog.csdn.net/Pig_cfbsl 
	> Created Time: 2020/5/2 20:44:09
 ************************************************************************/

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define For(i,a,b) for(register int i=(a);i<=(b);++i)
#define LL long long
using namespace std;
int read(){
	char x=getchar(); int u=0,fg=0;
	while(!isdigit(x)){ if(x=='-') fg=1; x=getchar(); }
	while(isdigit(x)){ u=(u<<3)+(u<<1)+(x^48); x=getchar(); }
	return fg?-u:u;
}
LL a,b,n;
int main()
{
#ifndef ONLINE_JUDGE
	freopen("input.in", "r", stdin);
	freopen("output.out", "w", stdout);
#endif
	scanf("%lld%lld%lld",&a,&b,&n);
	if(n>=b-1) printf("%lld\n",a*(b-1)/b);
	else printf("%lld\n",a*n/b);
	return 0;
}