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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define int long long 
#define f(i,x,n) for(int i=x;i<n;i++) 
#define all(c) c.begin(),c.end()
const int MOD = 1e9 + 7, N = 5e5 + 10;
const int LINF = LLONG_MAX;

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    double a,b,n;
    cin >> a >> b >> n;
    int ans = 0;
    f(i,1,min(a+1ll,n+1)){
        int r = floor(a*i/b);
        r = r-a*floor(i/b);
        ans = max(ans,r);
    }   
    cout << ans << "\n";
    return 0;
}