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)

#include <bits/stdc++.h>
#define ll long long
#define ld long double
#define fi first
#define se second
#define pb push_back
#define pii pair<int,int>
#define pll pair<long long,long long>
using namespace std;

ll a,b,n,x;

int main()
{

    ios_base::sync_with_stdio(false);
    cin.tie(0);

    cin >> a >> b >> n;
    if( n < b ) x = n;
    if( n >= b ) x = max( 1LL , b-1 );

    cout << ((a*x)/b)-a*(x/b);

    return 0;
}