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;

typedef long long ll;
ll mod = 1000000007;

ll power(ll x, ll y)
{
    ll res = 1;
    x = x%mod;
    while (y)
    {
        if (y&1) res = ((res%mod)*(x%mod))%mod;

        y>>=1;
        x = ((x%mod)*(x%mod))%mod;
    }
    return res;
}

int main(){

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

    ll n, a, b;

    cin >> a >> b >> n;

    ll ans;

    // if(a>b || n>b){
    //     if(a>b) ans =  n*floor(a/b)-(a*floor(n/b));
    //     else ans =  a*floor(n/b)-(a*floor(n/b));
    // }else{
    //     ans =  floor((a*n)/b)-(a*floor(n/b));
    // }

    ans = floor((n*a)/b)-a*floor(n/b);

    cout<<ans<<'\n';
    return 0;
}