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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;
using P = pair<int,int>;
const int MOD = 1000000007;

ll func(ll a,ll b,ll x) {
    return floor(a*x/b)-a*floor(x/b);
}

int main() {
    ll a,b,n;
    cin>>a>>b>>n;
    ll left=0;
    ll right=n+1;
    ll mid;
    ll maxi=func(a,b,1);
    while(right-left>1) {
        mid=(left+right)/2;
        if (func(a,b,mid)>maxi) {
            left=mid;
            maxi=func(a,b,mid);
        }
        else right=mid;
    }
    cout<<maxi<<endl;
}